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CyC2018
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@ -3371,6 +3371,7 @@ class MyQueue {
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```java
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class MyQueue {
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private Stack<Integer> in = new Stack();
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private Stack<Integer> out = new Stack();
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@ -3524,8 +3525,8 @@ public int[] dailyTemperatures(int[] temperatures) {
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int n = temperatures.length;
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int[] ret = new int[n];
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Stack<Integer> stack = new Stack<>();
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for(int i = 0; i < n; i++) {
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while(!stack.isEmpty() && temperatures[i] > temperatures[stack.peek()]) {
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for (int i = 0; i < n; i++) {
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while (!stack.isEmpty() && temperatures[i] > temperatures[stack.peek()]) {
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int idx = stack.pop();
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ret[idx] = i - idx;
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}
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@ -3548,15 +3549,15 @@ Output: [-1,3,-1]
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public int[] nextGreaterElement(int[] nums1, int[] nums2) {
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Map<Integer, Integer> map = new HashMap<>();
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Stack<Integer> stack = new Stack<>();
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for(int num : nums2){
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while(!stack.isEmpty() && num > stack.peek()){
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for (int num : nums2) {
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while (!stack.isEmpty() && num > stack.peek()) {
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map.put(stack.pop(), num);
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}
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stack.add(num);
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}
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int[] ret = new int[nums1.length];
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for(int i = 0; i < nums1.length; i++){
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if(map.containsKey(nums1[i])) ret[i] = map.get(nums1[i]);
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for (int i = 0; i < nums1.length; i++) {
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if (map.containsKey(nums1[i])) ret[i] = map.get(nums1[i]);
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else ret[i] = -1;
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}
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return ret;
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@ -3592,7 +3593,7 @@ Java 中的 **HashSet** 用于存储一个集合，并以 O(1) 的时间复杂
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Java 中的 **HashMap** 主要用于映射关系，从而把两个元素联系起来。
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在对一个内容进行压缩或者其它转换时，利用 HashMap 可以把原始内容和转换后的内容联系起来。例如在一个简化 url 的系统中（[Leetcdoe : 535. Encode and Decode TinyURL (Medium)](https://leetcode.com/problems/encode-and-decode-tinyurl/description/)），利用 HashMap 就可以存储精简后的 url 到原始 url 的映射，使得不仅可以显示简化的 url，也可以根据简化的 url 得到原始 url 从而定位到正确的资源。
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在对一个内容进行压缩或者其它转换时，利用 HashMap 可以把原始内容和转换后的内容联系起来。例如在一个简化 url 的系统中[Leetcdoe : 535. Encode and Decode TinyURL (Medium)](https://leetcode.com/problems/encode-and-decode-tinyurl/description/)，利用 HashMap 就可以存储精简后的 url 到原始 url 的映射，使得不仅可以显示简化的 url，也可以根据简化的 url 得到原始 url 从而定位到正确的资源。
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HashMap 也可以用来对元素进行计数统计，此时键为元素，值为计数。和 HashSet 类似，如果元素有穷并且范围不大，可以用整型数组来进行统计。
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@ -3602,13 +3603,13 @@ HashMap 也可以用来对元素进行计数统计，此时键为元素，值为
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可以先对数组进行排序，然后使用双指针方法或者二分查找方法。这样做的时间复杂度为 O(NlogN)，空间复杂度为 O(1)。
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用 HashMap 存储数组元素和索引的映射，在访问到 nums[i] 时，判断 HashMap 中是否存在 target - nums[i] ，如果存在说明 target - nums[i] 所在的索引和 i 就是要找的两个数。该方法的时间复杂度为 O(N)，空间复杂度为 O(N)，使用空间来换取时间。
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用 HashMap 存储数组元素和索引的映射，在访问到 nums[i] 时，判断 HashMap 中是否存在 target - nums[i]，如果存在说明 target - nums[i] 所在的索引和 i 就是要找的两个数。该方法的时间复杂度为 O(N)，空间复杂度为 O(N)，使用空间来换取时间。
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```java
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public int[] twoSum(int[] nums, int target) {
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HashMap<Integer, Integer> map = new HashMap<>();
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for(int i = 0; i < nums.length; i++){
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if(map.containsKey(target - nums[i])) return new int[]{map.get(target - nums[i]), i};
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for (int i = 0; i < nums.length; i++) {
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if (map.containsKey(target - nums[i])) return new int[] { map.get(target - nums[i]), i };
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else map.put(nums[i], i);
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}
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return null;
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@ -3622,7 +3623,9 @@ public int[] twoSum(int[] nums, int target) {
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```java
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public boolean containsDuplicate(int[] nums) {
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Set<Integer> set = new HashSet<>();
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for (int num : nums) set.add(num);
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for (int num : nums) {
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set.add(num);
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}
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return set.size() < nums.length;
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}
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```
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@ -3657,13 +3660,15 @@ public int findLHS(int[] nums) {
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**最长连续序列**
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[Leetcode : 128. Longest Consecutive Sequence (Medium)](https://leetcode.com/problems/longest-consecutive-sequence/description/)
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[Leetcode : 128. Longest Consecutive Sequence (Hard)](https://leetcode.com/problems/longest-consecutive-sequence/description/)
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```html
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Given [100, 4, 200, 1, 3, 2],
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The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.
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```
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题目要求：以 O(N) 的时间复杂度求解。
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```java
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public int longestConsecutive(int[] nums) {
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Map<Integer, Integer> numCnts = new HashMap<>();
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@ -3705,41 +3710,14 @@ s = "anagram", t = "nagaram", return true.
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s = "rat", t = "car", return false.
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```
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字符串只包含小写字符，总共有 26 个小写字符。可以用 Hash Table 来映射字符与出现次数，因为键值范围很小，因此可以使用长度为 26 的整型数组对字符串出现的字符进行统计，比较两个字符串出现的字符数量是否相同。
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字符串只包含小写字符，总共有 26 个小写字符。可以用 Hash Table 来映射字符与出现次数，因为键值范围很小，因此可以使用长度为 26 的整型数组对字符串出现的字符进行统计，然后比较两个字符串出现的字符数量是否相同。
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```java
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public boolean isAnagram(String s, String t) {
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int[] cnts = new int[26];
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for(int i = 0; i < s.length(); i++) cnts[s.charAt(i) - 'a']++;
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for(int i = 0; i < t.length(); i++) cnts[t.charAt(i) - 'a']--;
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for(int i = 0; i < 26; i++) if(cnts[i] != 0) return false;
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return true;
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}
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```
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**字符串同构**
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[Leetcode : 205. Isomorphic Strings (Easy)](https://leetcode.com/problems/isomorphic-strings/description/)
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```html
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Given "egg", "add", return true.
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Given "foo", "bar", return false.
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Given "paper", "title", return true.
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```
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记录一个字符上次出现的位置，如果两个字符串中某个字符上次出现的位置一样，那么就属于同构。
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```java
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public boolean isIsomorphic(String s, String t) {
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int[] m1 = new int[256];
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int[] m2 = new int[256];
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for(int i = 0; i < s.length(); i++){
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if(m1[s.charAt(i)] != m2[t.charAt(i)]) {
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return false;
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}
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m1[s.charAt(i)] = i + 1;
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m2[t.charAt(i)] = i + 1;
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}
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for (char c : s.toCharArray()) cnts[c - 'a']++;
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for (char c : t.toCharArray()) cnts[c - 'a']--;
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for (int cnt : cnts) if (cnt != 0) return false;
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return true;
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}
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```
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@ -3754,34 +3732,60 @@ Output : 7
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Explanation : One longest palindrome that can be built is "dccaccd", whose length is 7.
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```
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使用长度为 128 的整型数组来统计每个字符出现的个数，每个字符有偶数个可以用来构成回文字符串。因为回文字符串最中间的那个字符可以单独出现，所以如果有单独的字符就把它放到最中间。
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使用长度为 256 的整型数组来统计每个字符出现的个数，每个字符有偶数个可以用来构成回文字符串。因为回文字符串最中间的那个字符可以单独出现，所以如果有单独的字符就把它放到最中间。
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```java
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public int longestPalindrome(String s) {
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int[] cnts = new int[128]; // ascii 码总共 128 个
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for(char c : s.toCharArray()) cnts[c]++;
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int[] cnts = new int[256];
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for (char c : s.toCharArray()) cnts[c]++;
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int ret = 0;
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for(int cnt : cnts) ret += (cnt / 2) * 2;
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if(ret < s.length()) ret++; // 这个条件下 s 中一定有单个未使用的字符存在，可以把这个字符放到回文的最中间
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for (int cnt : cnts) ret += (cnt / 2) * 2;
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if (ret < s.length()) ret++; // 这个条件下 s 中一定有单个未使用的字符存在，可以把这个字符放到回文的最中间
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return ret;
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}
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```
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**字符串同构**
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[Leetcode : 205. Isomorphic Strings (Easy)](https://leetcode.com/problems/isomorphic-strings/description/)
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```html
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Given "egg", "add", return true.
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Given "foo", "bar", return false.
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Given "paper", "title", return true.
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```
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记录一个字符上次出现的位置，如果两个字符串中的字符上次出现的位置一样，那么就属于同构。
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```java
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public boolean isIsomorphic(String s, String t) {
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int[] preIndexOfS = new int[256];
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int[] preIndexOfT = new int[256];
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for (int i = 0; i < s.length(); i++) {
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char sc = s.charAt(i), tc = t.charAt(i);
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if (preIndexOfS[sc] != preIndexOfT[tc]) return false;
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preIndexOfS[sc] = i + 1;
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preIndexOfT[tc] = i + 1;
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}
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return true;
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}
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```
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**判断一个整数是否是回文数**
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[Leetcode : 9. Palindrome Number (Easy)](https://leetcode.com/problems/palindrome-number/description/)
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要求不能使用额外空间，也就不能将整数转换为字符串进行判断。
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题目要求：不能使用额外空间，也就不能将整数转换为字符串进行判断。
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将整数分成左右两部分，右边那部分需要转置，然后判断这两部分是否相等。
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```java
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public boolean isPalindrome(int x) {
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if(x == 0) return true;
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if(x < 0) return false;
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if(x % 10 == 0) return false;
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if (x == 0) return true;
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if (x < 0) return false;
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if (x % 10 == 0) return false;
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int right = 0;
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while(x > right){
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while (x > right) {
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right = right * 10 + x % 10;
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x /= 10;
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}
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@ -3804,7 +3808,7 @@ Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
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```java
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private int cnt = 0;
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public int countSubstrings(String s) {
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for(int i = 0; i < s.length(); i++) {
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for (int i = 0; i < s.length(); i++) {
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extendSubstrings(s, i, i); // 奇数长度
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extendSubstrings(s, i, i + 1); // 偶数长度
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}
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@ -3812,7 +3816,7 @@ public int countSubstrings(String s) {
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}
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private void extendSubstrings(String s, int start, int end) {
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while(start >= 0 && end < s.length() && s.charAt(start) == s.charAt(end)) {
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while (start >= 0 && end < s.length() && s.charAt(start) == s.charAt(end)) {
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start--;
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end++;
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cnt++;
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@ -3833,14 +3837,13 @@ Explanation: There are 6 substrings that have equal number of consecutive 1's an
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```java
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public int countBinarySubstrings(String s) {
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int preLen = 0, curLen = 1, ret = 0;
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for(int i = 1; i < s.length(); i++){
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if(s.charAt(i) == s.charAt(i-1)) curLen++;
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else{
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for (int i = 1; i < s.length(); i++) {
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if (s.charAt(i) == s.charAt(i-1)) curLen++;
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else {
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preLen = curLen;
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curLen = 1;
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}
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if(preLen >= curLen) ret++;
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if (preLen >= curLen) ret++;
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}
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return ret;
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}
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@ -3848,7 +3851,7 @@ public int countBinarySubstrings(String s) {
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**字符串循环移位包含**
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[ 编程之美：3.1](#)
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[编程之美：3.1](#)
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```html
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s1 = AABCD, s2 = CDAA
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@ -3861,7 +3864,7 @@ s1 进行循环移位的结果是 s1s1 的子字符串，因此只要判断 s2
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**字符串循环移位**
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[ 编程之美：2.17](#)
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[编程之美：2.17](#)
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将字符串向右循环移动 k 位。
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@ -3900,12 +3903,12 @@ public void moveZeroes(int[] nums) {
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[Leetcode : 566. Reshape the Matrix (Easy)](https://leetcode.com/problems/reshape-the-matrix/description/)
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```html
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Input:
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nums =
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Input:
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nums =
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[[1,2],
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[3,4]]
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r = 1, c = 4
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Output:
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Output:
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[[1,2,3,4]]
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Explanation:
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The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.
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@ -3933,19 +3936,191 @@ public int[][] matrixReshape(int[][] nums, int r, int c) {
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```java
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public int findMaxConsecutiveOnes(int[] nums) {
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int max = 0;
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int cur = 0;
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int max = 0, cur = 0;
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for (int num : nums) {
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if (num == 0) cur = 0;
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else {
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cur++;
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max = Math.max(max, cur);
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}
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cur = num == 0 ? 0 : cur + 1;
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max = Math.max(max, cur);
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}
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return max;
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}
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```
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**一个数组元素在 [1, n] 之间，其中一个数被替换为另一个数，找出丢失的数和重复的数**
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[Leetcode : 645. Set Mismatch (Easy)](https://leetcode.com/problems/set-mismatch/description/)
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```html
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Input: nums = [1,2,2,4]
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Output: [2,3]
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```
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```html
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Input: nums = [1,2,2,4]
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Output: [2,3]
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```
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最直接的方法是先对数组进行排序，这种方法时间复杂度为 O(NlogN)。本题可以以 O(N) 的时间复杂度、O(1) 空间复杂度来求解。
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主要思想是通过交换数组元素，使得数组上的元素在正确的位置上。遍历数组，如果第 i 位上的元素不是 i + 1，那么就交换第 i 位和 nums[i] - 1 位上的元素，使得 num[i] - 1 位置上的元素为 nums[i]，也就是该位置上的元素是正确的。
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```java
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public int[] findErrorNums(int[] nums) {
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for (int i = 0; i < nums.length; i++) {
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while (nums[i] != i + 1) {
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if (nums[i] == nums[nums[i] - 1]) {
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return new int[]{nums[nums[i] - 1], i + 1};
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}
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swap(nums, i, nums[i] - 1);
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}
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}
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return null;
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}
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private void swap(int[] nums, int i, int j) {
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int tmp = nums[i]; nums[i] = nums[j]; nums[j] = tmp;
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}
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```
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类似题目：
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- [Leetcode :448. Find All Numbers Disappeared in an Array (Easy)](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/description/)，寻找所有丢失的元素
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- [Leetcode : 442. Find All Duplicates in an Array (Medium)](https://leetcode.com/problems/find-all-duplicates-in-an-array/description/)，寻找所有重复的元素。
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**找出数组中重复的数，数组值在 [1, n] 之间**
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[Leetcode : 287. Find the Duplicate Number (Medium)](https://leetcode.com/problems/find-the-duplicate-number/description/)
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要求不能修改数组，也不能使用额外的空间。
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二分查找解法：
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|
||||
```java
|
||||
public int findDuplicate(int[] nums) {
|
||||
int l = 1, h = nums.length - 1;
|
||||
while (l <= h) {
|
||||
int mid = l + (h - l) / 2;
|
||||
int cnt = 0;
|
||||
for (int i = 0; i < nums.length; i++) {
|
||||
if (nums[i] <= mid) cnt++;
|
||||
}
|
||||
if (cnt > mid) h = mid - 1;
|
||||
else l = mid + 1;
|
||||
}
|
||||
return l;
|
||||
}
|
||||
```
|
||||
|
||||
双指针解法，类似于有环链表中找出环的入口：
|
||||
|
||||
```java
|
||||
public int findDuplicate(int[] nums) {
|
||||
int slow = nums[0], fast = nums[nums[0]];
|
||||
while (slow != fast) {
|
||||
slow = nums[slow];
|
||||
fast = nums[nums[fast]];
|
||||
}
|
||||
fast = 0;
|
||||
while (slow != fast) {
|
||||
slow = nums[slow];
|
||||
fast = nums[fast];
|
||||
}
|
||||
return slow;
|
||||
}
|
||||
```
|
||||
|
||||
**有序矩阵查找**
|
||||
|
||||
[Leetocde : 240. Search a 2D Matrix II (Medium)](https://leetcode.com/problems/search-a-2d-matrix-ii/description/)
|
||||
|
||||
```html
|
||||
[
|
||||
[ 1, 5, 9],
|
||||
[10, 11, 13],
|
||||
[12, 13, 15]
|
||||
]
|
||||
```
|
||||
|
||||
```java
|
||||
public boolean searchMatrix(int[][] matrix, int target) {
|
||||
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
|
||||
int m = matrix.length, n = matrix[0].length;
|
||||
int row = 0, col = n - 1;
|
||||
while (row < m && col >= 0) {
|
||||
if (target == matrix[row][col]) return true;
|
||||
else if (target < matrix[row][col]) col--;
|
||||
else row++;
|
||||
}
|
||||
return false;
|
||||
}
|
||||
```
|
||||
|
||||
**有序矩阵的 Kth Element**
|
||||
|
||||
[Leetcode : 378. Kth Smallest Element in a Sorted Matrix ((Medium))](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/description/)
|
||||
|
||||
```html
|
||||
matrix = [
|
||||
[ 1, 5, 9],
|
||||
[10, 11, 13],
|
||||
[12, 13, 15]
|
||||
],
|
||||
k = 8,
|
||||
|
||||
return 13.
|
||||
```
|
||||
|
||||
解题参考：[Share my thoughts and Clean Java Code](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/85173)
|
||||
|
||||
二分查找解法：
|
||||
|
||||
```java
|
||||
public int kthSmallest(int[][] matrix, int k) {
|
||||
int m = matrix.length, n = matrix[0].length;
|
||||
int lo = matrix[0][0], hi = matrix[m - 1][n - 1];
|
||||
while(lo <= hi) {
|
||||
int mid = lo + (hi - lo) / 2;
|
||||
int cnt = 0;
|
||||
for(int i = 0; i < m; i++) {
|
||||
for(int j = 0; j < n && matrix[i][j] <= mid; j++) {
|
||||
cnt++;
|
||||
}
|
||||
}
|
||||
if(cnt < k) lo = mid + 1;
|
||||
else hi = mid - 1;
|
||||
}
|
||||
return lo;
|
||||
}
|
||||
```
|
||||
|
||||
堆解法：
|
||||
|
||||
```java
|
||||
public int kthSmallest(int[][] matrix, int k) {
|
||||
int m = matrix.length, n = matrix[0].length;
|
||||
PriorityQueue<Tuple> pq = new PriorityQueue<Tuple>();
|
||||
for(int j = 0; j < n; j++) pq.offer(new Tuple(0, j, matrix[0][j]));
|
||||
for(int i = 0; i < k - 1; i++) { // 小根堆，去掉 k - 1 个堆顶元素，此时堆顶元素就是第 k 的数
|
||||
Tuple t = pq.poll();
|
||||
if(t.x == m - 1) continue;
|
||||
pq.offer(new Tuple(t.x + 1, t.y, matrix[t.x + 1][t.y]));
|
||||
}
|
||||
return pq.poll().val;
|
||||
}
|
||||
|
||||
class Tuple implements Comparable<Tuple> {
|
||||
int x, y, val;
|
||||
public Tuple(int x, int y, int val) {
|
||||
this.x = x; this.y = y; this.val = val;
|
||||
}
|
||||
|
||||
@Override
|
||||
public int compareTo(Tuple that) {
|
||||
return this.val - that.val;
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
**数组相邻差值的个数**
|
||||
|
||||
[Leetcode : 667. Beautiful Arrangement II (Medium)](https://leetcode.com/problems/beautiful-arrangement-ii/description/)
|
||||
|
|
@ -4112,185 +4287,6 @@ public int maxChunksToSorted(int[] arr) {
|
|||
}
|
||||
```
|
||||
|
||||
|
||||
**一个数组元素在 [1, n] 之间，其中一个数被替换为另一个数，找出丢失的数和重复的数**
|
||||
|
||||
[Leetcode : 645. Set Mismatch (Easy)](https://leetcode.com/problems/set-mismatch/description/)
|
||||
|
||||
```html
|
||||
Input: nums = [1,2,2,4]
|
||||
Output: [2,3]
|
||||
```
|
||||
|
||||
```html
|
||||
Input: nums = [1,2,2,4]
|
||||
Output: [2,3]
|
||||
```
|
||||
|
||||
最直接的方法是先对数组进行排序，这种方法时间复杂度为 O(nlogn)。本题可以以 O(n) 的时间复杂度、O(1) 空间复杂度来求解。
|
||||
|
||||
主要思想是通过交换数组元素，使得数组上的元素在正确的位置上。
|
||||
|
||||
遍历数组，如果第 i 位上的元素不是 i + 1 ，那么就交换第 i 位 和 nums[i] - 1 位上的元素，使得 num[i] - 1 的元素为 nums[i] ，也就是该位的元素是正确的。交换操作需要循环进行，因为一次交换没办法使得第 i 位上的元素是正确的。但是要交换的两个元素可能就是重复元素，那么循环就可能永远进行下去，终止循环的方法是加上 nums[i] != nums[nums[i] - 1 条件。
|
||||
|
||||
类似题目：
|
||||
|
||||
- [Leetcode :448. Find All Numbers Disappeared in an Array (Easy)](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/description/)，寻找所有丢失的元素
|
||||
- [Leetcode : 442. Find All Duplicates in an Array (Medium)](https://leetcode.com/problems/find-all-duplicates-in-an-array/description/)，寻找所有重复的元素。
|
||||
|
||||
```java
|
||||
public int[] findErrorNums(int[] nums) {
|
||||
for (int i = 0; i < nums.length; i++) {
|
||||
while (nums[i] != i + 1) {
|
||||
if (nums[i] == nums[nums[i] - 1]) {
|
||||
return new int[]{nums[nums[i] - 1], i + 1};
|
||||
}
|
||||
swap(nums, i, nums[i] - 1);
|
||||
}
|
||||
}
|
||||
|
||||
return null;
|
||||
}
|
||||
|
||||
private void swap(int[] nums, int i, int j) {
|
||||
int tmp = nums[i]; nums[i] = nums[j]; nums[j] = tmp;
|
||||
}
|
||||
```
|
||||
|
||||
**找出数组中重复的数，数组值在 [1, n] 之间**
|
||||
|
||||
[Leetcode : 287. Find the Duplicate Number (Medium)](https://leetcode.com/problems/find-the-duplicate-number/description/)
|
||||
|
||||
要求不能修改数组，也不能使用额外的空间。
|
||||
|
||||
二分查找解法：
|
||||
|
||||
```java
|
||||
public int findDuplicate(int[] nums) {
|
||||
int l = 1, h = nums.length - 1;
|
||||
while (l <= h) {
|
||||
int mid = l + (h - l) / 2;
|
||||
int cnt = 0;
|
||||
for (int i = 0; i < nums.length; i++) {
|
||||
if (nums[i] <= mid) cnt++;
|
||||
}
|
||||
if (cnt > mid) h = mid - 1;
|
||||
else l = mid + 1;
|
||||
}
|
||||
return l;
|
||||
}
|
||||
```
|
||||
|
||||
双指针解法，类似于有环链表中找出环的入口：
|
||||
|
||||
```java
|
||||
public int findDuplicate(int[] nums) {
|
||||
int slow = nums[0], fast = nums[nums[0]];
|
||||
while (slow != fast) {
|
||||
slow = nums[slow];
|
||||
fast = nums[nums[fast]];
|
||||
}
|
||||
fast = 0;
|
||||
while (slow != fast) {
|
||||
slow = nums[slow];
|
||||
fast = nums[fast];
|
||||
}
|
||||
return slow;
|
||||
}
|
||||
```
|
||||
|
||||
**有序矩阵查找**
|
||||
|
||||
[Leetocde : 240. Search a 2D Matrix II (Medium)](https://leetcode.com/problems/search-a-2d-matrix-ii/description/)
|
||||
|
||||
```html
|
||||
[
|
||||
[ 1, 5, 9],
|
||||
[10, 11, 13],
|
||||
[12, 13, 15]
|
||||
]
|
||||
```
|
||||
|
||||
```java
|
||||
public boolean searchMatrix(int[][] matrix, int target) {
|
||||
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
|
||||
int m = matrix.length, n = matrix[0].length;
|
||||
int row = 0, col = n - 1;
|
||||
while (row < m && col >= 0) {
|
||||
if (target == matrix[row][col]) return true;
|
||||
else if (target < matrix[row][col]) col--;
|
||||
else row++;
|
||||
}
|
||||
return false;
|
||||
}
|
||||
```
|
||||
|
||||
**有序矩阵的 Kth Element**
|
||||
|
||||
[Leetcode : 378. Kth Smallest Element in a Sorted Matrix ((Medium))](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/description/)
|
||||
|
||||
```html
|
||||
matrix = [
|
||||
[ 1, 5, 9],
|
||||
[10, 11, 13],
|
||||
[12, 13, 15]
|
||||
],
|
||||
k = 8,
|
||||
|
||||
return 13.
|
||||
```
|
||||
|
||||
解题参考：[Share my thoughts and Clean Java Code](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/85173)
|
||||
|
||||
二分查找解法：
|
||||
|
||||
```java
|
||||
public int kthSmallest(int[][] matrix, int k) {
|
||||
int m = matrix.length, n = matrix[0].length;
|
||||
int lo = matrix[0][0], hi = matrix[m - 1][n - 1];
|
||||
while(lo <= hi) {
|
||||
int mid = lo + (hi - lo) / 2;
|
||||
int cnt = 0;
|
||||
for(int i = 0; i < m; i++) {
|
||||
for(int j = 0; j < n && matrix[i][j] <= mid; j++) {
|
||||
cnt++;
|
||||
}
|
||||
}
|
||||
if(cnt < k) lo = mid + 1;
|
||||
else hi = mid - 1;
|
||||
}
|
||||
return lo;
|
||||
}
|
||||
```
|
||||
|
||||
堆解法：
|
||||
|
||||
```java
|
||||
public int kthSmallest(int[][] matrix, int k) {
|
||||
int m = matrix.length, n = matrix[0].length;
|
||||
PriorityQueue<Tuple> pq = new PriorityQueue<Tuple>();
|
||||
for(int j = 0; j < n; j++) pq.offer(new Tuple(0, j, matrix[0][j]));
|
||||
for(int i = 0; i < k - 1; i++) { // 小根堆，去掉 k - 1 个堆顶元素，此时堆顶元素就是第 k 的数
|
||||
Tuple t = pq.poll();
|
||||
if(t.x == m - 1) continue;
|
||||
pq.offer(new Tuple(t.x + 1, t.y, matrix[t.x + 1][t.y]));
|
||||
}
|
||||
return pq.poll().val;
|
||||
}
|
||||
|
||||
class Tuple implements Comparable<Tuple> {
|
||||
int x, y, val;
|
||||
public Tuple(int x, int y, int val) {
|
||||
this.x = x; this.y = y; this.val = val;
|
||||
}
|
||||
|
||||
@Override
|
||||
public int compareTo(Tuple that) {
|
||||
return this.val - that.val;
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
## 链表
|
||||
|
||||
链表是空节点，或者有一个值和一个指向下一个链表的指针，因此很多链表问题可以用递归来处理。
|
||||
|
|
@ -4316,7 +4312,7 @@ B: b1 → b2 → b3
|
|||
```java
|
||||
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
|
||||
ListNode l1 = headA, l2 = headB;
|
||||
while(l1 != l2){
|
||||
while (l1 != l2) {
|
||||
l1 = (l1 == null) ? headB : l1.next;
|
||||
l2 = (l2 == null) ? headA : l2.next;
|
||||
}
|
||||
|
|
@ -4447,52 +4443,6 @@ public ListNode swapPairs(ListNode head) {
|
|||
}
|
||||
```
|
||||
|
||||
**根据有序链表构造平衡的 BST**
|
||||
|
||||
[Leetcode : 109. Convert Sorted List to Binary Search Tree (Medium)](https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/description/)
|
||||
|
||||
```html
|
||||
Given the sorted linked list: [-10,-3,0,5,9],
|
||||
|
||||
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
|
||||
|
||||
0
|
||||
/ \
|
||||
-3 9
|
||||
/ /
|
||||
-10 5
|
||||
```
|
||||
|
||||
```java
|
||||
public TreeNode sortedListToBST(ListNode head) {
|
||||
if (head == null) return null;
|
||||
int size = size(head);
|
||||
if (size == 1) return new TreeNode(head.val);
|
||||
ListNode pre = head, mid = pre.next;
|
||||
int step = 2;
|
||||
while (step <= size / 2) {
|
||||
pre = mid;
|
||||
mid = mid.next;
|
||||
step++;
|
||||
}
|
||||
pre.next = null;
|
||||
TreeNode t = new TreeNode(mid.val);
|
||||
t.left = sortedListToBST(head);
|
||||
t.right = sortedListToBST(mid.next);
|
||||
return t;
|
||||
}
|
||||
|
||||
private int size(ListNode node) {
|
||||
int size = 0;
|
||||
while (node != null) {
|
||||
size++;
|
||||
node = node.next;
|
||||
}
|
||||
return size;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
**链表求和**
|
||||
|
||||
[Leetcode : 445. Add Two Numbers II (Medium)](https://leetcode.com/problems/add-two-numbers-ii/description/)
|
||||
|
|
@ -4532,46 +4482,6 @@ private Stack<Integer> buildStack(ListNode l) {
|
|||
}
|
||||
```
|
||||
|
||||
**分隔链表**
|
||||
|
||||
[Leetcode : 725. Split Linked List in Parts(Medium)](https://leetcode.com/problems/split-linked-list-in-parts/description/)
|
||||
|
||||
```html
|
||||
Input:
|
||||
root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
|
||||
Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
|
||||
Explanation:
|
||||
The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.
|
||||
```
|
||||
|
||||
题目描述：把链表分隔成 k 部分，每部分的长度都应该尽可能相同，排在前面的长度应该大于等于后面的。
|
||||
|
||||
```java
|
||||
public ListNode[] splitListToParts(ListNode root, int k) {
|
||||
int N = 0;
|
||||
ListNode cur = root;
|
||||
while (cur != null) {
|
||||
N++;
|
||||
cur = cur.next;
|
||||
}
|
||||
int mod = N % k;
|
||||
int size = N / k;
|
||||
ListNode[] ret = new ListNode[k];
|
||||
cur = root;
|
||||
for (int i = 0; cur != null && i < k; i++) {
|
||||
ret[i] = cur;
|
||||
int curSize = size + (mod-- > 0 ? 1 : 0);
|
||||
for (int j = 0; j < curSize - 1; j++) {
|
||||
cur = cur.next;
|
||||
}
|
||||
ListNode next = cur.next;
|
||||
cur.next = null;
|
||||
cur = next;
|
||||
}
|
||||
return ret;
|
||||
}
|
||||
```
|
||||
|
||||
**回文链表**
|
||||
|
||||
[Leetcode : 234. Palindrome Linked List (Easy)](https://leetcode.com/problems/palindrome-linked-list/description/)
|
||||
|
|
@ -4652,6 +4562,46 @@ public ListNode oddEvenList(ListNode head) {
|
|||
}
|
||||
```
|
||||
|
||||
**分隔链表**
|
||||
|
||||
[Leetcode : 725. Split Linked List in Parts(Medium)](https://leetcode.com/problems/split-linked-list-in-parts/description/)
|
||||
|
||||
```html
|
||||
Input:
|
||||
root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
|
||||
Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
|
||||
Explanation:
|
||||
The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.
|
||||
```
|
||||
|
||||
题目描述：把链表分隔成 k 部分，每部分的长度都应该尽可能相同，排在前面的长度应该大于等于后面的。
|
||||
|
||||
```java
|
||||
public ListNode[] splitListToParts(ListNode root, int k) {
|
||||
int N = 0;
|
||||
ListNode cur = root;
|
||||
while (cur != null) {
|
||||
N++;
|
||||
cur = cur.next;
|
||||
}
|
||||
int mod = N % k;
|
||||
int size = N / k;
|
||||
ListNode[] ret = new ListNode[k];
|
||||
cur = root;
|
||||
for (int i = 0; cur != null && i < k; i++) {
|
||||
ret[i] = cur;
|
||||
int curSize = size + (mod-- > 0 ? 1 : 0);
|
||||
for (int j = 0; j < curSize - 1; j++) {
|
||||
cur = cur.next;
|
||||
}
|
||||
ListNode next = cur.next;
|
||||
cur.next = null;
|
||||
cur = next;
|
||||
}
|
||||
return ret;
|
||||
}
|
||||
```
|
||||
|
||||
## 树
|
||||
|
||||
### 递归
|
||||
|
|
@ -5545,6 +5495,52 @@ private void inOrder(TreeNode node, int k) {
|
|||
}
|
||||
```
|
||||
|
||||
**根据有序链表构造平衡的 BST**
|
||||
|
||||
[Leetcode : 109. Convert Sorted List to Binary Search Tree (Medium)](https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/description/)
|
||||
|
||||
```html
|
||||
Given the sorted linked list: [-10,-3,0,5,9],
|
||||
|
||||
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
|
||||
|
||||
0
|
||||
/ \
|
||||
-3 9
|
||||
/ /
|
||||
-10 5
|
||||
```
|
||||
|
||||
```java
|
||||
public TreeNode sortedListToBST(ListNode head) {
|
||||
if (head == null) return null;
|
||||
int size = size(head);
|
||||
if (size == 1) return new TreeNode(head.val);
|
||||
ListNode pre = head, mid = pre.next;
|
||||
int step = 2;
|
||||
while (step <= size / 2) {
|
||||
pre = mid;
|
||||
mid = mid.next;
|
||||
step++;
|
||||
}
|
||||
pre.next = null;
|
||||
TreeNode t = new TreeNode(mid.val);
|
||||
t.left = sortedListToBST(head);
|
||||
t.right = sortedListToBST(mid.next);
|
||||
return t;
|
||||
}
|
||||
|
||||
private int size(ListNode node) {
|
||||
int size = 0;
|
||||
while (node != null) {
|
||||
size++;
|
||||
node = node.next;
|
||||
}
|
||||
return size;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
### Trie
|
||||
|
||||
<div align="center"> <img src="../pics//5c638d59-d4ae-4ba4-ad44-80bdc30f38dd.jpg"/> </div><br>
|
||||
|
|
|
|||
Loading…
Reference in New Issue
Block a user