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CyC2018
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@ -263,7 +263,7 @@ Output: 1
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```java
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public int findMin(int[] nums) {
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int l = 0, h = nums.length;
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int l = 0, h = nums.length - 1;
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while (l < h) {
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int m = l + (h - l) / 2;
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if (nums[m] <= nums[h]) h = m;
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@ -909,15 +909,15 @@ public ListNode deleteNode(ListNode head, ListNode tobeDelete) {
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```java
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public ListNode deleteDuplication(ListNode pHead) {
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if (pHead == null) return null;
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if (pHead == null || pHead.next == null) return pHead;
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ListNode next = pHead.next;
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if (next == null) return pHead;
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if (pHead.val == next.val) {
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while (next != null && pHead.val == next.val) next = next.next;
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return deleteDuplication(next);
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} else {
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pHead.next = deleteDuplication(pHead.next);
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return pHead;
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}
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pHead.next = deleteDuplication(pHead.next);
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return pHead;
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}
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```
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@ -2256,6 +2256,8 @@ private int height(TreeNode root) {
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## 题目描述
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[NowCoder](https://www.nowcoder.com/practice/e02fdb54d7524710a7d664d082bb7811?tpId=13&tqId=11193&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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一个整型数组里除了两个数字之外，其他的数字都出现了两次，找出这两个数。
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## 解题思路
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@ -2267,40 +2269,49 @@ private int height(TreeNode root) {
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diff &= -diff 得到出 diff 最右侧不为 0 的位，也就是不存在重复的两个元素在位级表示上最右侧不同的那一位，利用这一位就可以将两个元素区分开来。
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```java
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public void FindNumsAppearOnce(int[] array, int num1[], int num2[]) {
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int diff = 0;
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for (int num : array) diff ^= num;
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// 得到最右一位
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diff &= -diff;
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for (int num : array) {
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if ((num & diff) == 0) num1[0] ^= num;
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else num2[0] ^= num;
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public void FindNumsAppearOnce(int[] nums, int num1[], int num2[]) {
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int diff = 0;
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for (int num : nums) {
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diff ^= num;
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}
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// 得到最右一位
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diff &= -diff;
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for (int num : nums) {
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if ((num & diff) == 0) {
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num1[0] ^= num;
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} else {
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num2[0] ^= num;
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}
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}
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}
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}
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```
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# 57.1 和为 S 的两个数字
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## 题目描述
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[NowCoder](https://www.nowcoder.com/practice/390da4f7a00f44bea7c2f3d19491311b?tpId=13&tqId=11195&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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输入一个递增排序的数组和一个数字 S，在数组中查找两个数，使得他们的和正好是 S，如果有多对数字的和等于 S，输出两个数的乘积最小的。
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## 解题思路
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使用双指针，一个指针指向元素较小的值，一个指针指向元素较大的值。指向较小元素的指针从头向尾遍历，指向较大元素的指针从尾向头遍历。
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如果两个指针指向元素的和 sum == target，那么得到要求的结果；如果 sum > target，移动较大的元素，使 sum 变小一些；如果 sum < target，移动较小的元素，使 sum 变大一些。
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- 如果两个指针指向元素的和 sum == target，那么得到要求的结果；
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- 如果 sum > target，移动较大的元素，使 sum 变小一些；
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- 如果 sum < target，移动较小的元素，使 sum 变大一些。
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```java
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public ArrayList<Integer> FindNumbersWithSum(int[] array, int sum) {
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int i = 0, j = array.length - 1;
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while (i < j) {
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int cur = array[i] + array[j];
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if (cur == sum) return new ArrayList<Integer>(Arrays.asList(array[i], array[j]));
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else if (cur < sum) i++;
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if (cur == sum) return new ArrayList<>(Arrays.asList(array[i], array[j]));
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if (cur < sum) i++;
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else j--;
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}
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return new ArrayList<Integer>();
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return new ArrayList<>();
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}
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```
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@ -2308,32 +2319,41 @@ public ArrayList<Integer> FindNumbersWithSum(int[] array, int sum) {
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## 题目描述
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和为 100 的连续序列有 18, 19, 20, 21, 22。
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[NowCoder](https://www.nowcoder.com/practice/c451a3fd84b64cb19485dad758a55ebe?tpId=13&tqId=11194&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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输出所有和为 S 的连续正数序列。序列内按照从小至大的顺序，序列间按照开始数字从小到大的顺序
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例如和为 100 的连续序列有：
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```
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[9, 10, 11, 12, 13, 14, 15, 16]
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[18, 19, 20, 21, 22]。
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```
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## 解题思路
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```java
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public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum) {
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ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
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int first = 1, last = 2;
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int start = 1, end = 2;
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int curSum = 3;
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while (first <= sum / 2 && last < sum) {
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while (end < sum) {
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if (curSum > sum) {
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curSum -= first;
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first++;
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curSum -= start;
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start++;
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} else if (curSum < sum) {
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last++;
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curSum += last;
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end++;
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curSum += end;
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} else {
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ArrayList<Integer> list = new ArrayList<>();
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for (int i = first; i <= last; i++) {
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for (int i = start; i <= end; i++) {
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list.add(i);
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}
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ret.add(list);
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curSum -= first;
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first++;
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last++;
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curSum += last;
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curSum -= start;
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start++;
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end++;
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curSum += end;
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}
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}
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return ret;
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@ -2350,11 +2370,14 @@ public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum) {
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## 解题思路
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题目应该有一个隐含条件，就是不能用额外的空间。虽然 Java 的题目输入参数为 String 类型，需要先创建一个字符数组使得空间复杂度为 O(n)，但是正确的参数类型应该和原书一样，为字符数组，并且只能使用该字符数组的空间。任何使用了额外空间的解法在面试时都会大打折扣，包括递归解法。正确的解法应该是和书上一样，先旋转每个单词，再旋转整个字符串。
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[NowCoder](https://www.nowcoder.com/practice/3194a4f4cf814f63919d0790578d51f3?tpId=13&tqId=11197&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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题目应该有一个隐含条件，就是不能用额外的空间。虽然 Java 的题目输入参数为 String 类型，需要先创建一个字符数组使得空间复杂度为 O(N)，但是正确的参数类型应该和原书一样，为字符数组，并且只能使用该字符数组的空间。任何使用了额外空间的解法在面试时都会大打折扣，包括递归解法。
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正确的解法应该是和书上一样，先旋转每个单词，再旋转整个字符串。
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```java
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public String ReverseSentence(String str) {
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if (str.length() == 0) return str;
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int n = str.length();
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char[] chars = str.toCharArray();
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int i = 0, j = 0;
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@ -2370,43 +2393,59 @@ public String ReverseSentence(String str) {
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}
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private void reverse(char[] c, int i, int j) {
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while(i < j) {
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char t = c[i]; c[i] = c[j]; c[j] = t;
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i++; j--;
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while (i < j) {
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swap(c, i++, j--);
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}
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}
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private void swap(char[] c, int i, int j) {
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char t = c[i];
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c[i] = c[j];
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c[j] = t;
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}
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```
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# 58.2 左旋转字符串
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## 题目描述
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[NowCoder](https://www.nowcoder.com/practice/12d959b108cb42b1ab72cef4d36af5ec?tpId=13&tqId=11196&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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对于一个给定的字符序列 S，请你把其循环左移 K 位后的序列输出。例如，字符序列 S=”abcXYZdef”, 要求输出循环左移 3 位后的结果，即“XYZdefabc”。
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## 解题思路
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将 "abcXYZdef" 旋转左移三位，可以先将 "abc" 和 "XYZdef" 分别旋转，得到 "cbafedZYX"，然后再把整个字符串旋转得到 "XYZdefabc"。
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```java
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public String LeftRotateString(String str, int n) {
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if(str.length() == 0) return "";
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char[] c = str.toCharArray();
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reverse(c, 0, n - 1);
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reverse(c, n, c.length - 1);
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reverse(c, 0, c.length - 1);
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return new String(c);
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if (n >= str.length()) return str;
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char[] chars = str.toCharArray();
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reverse(chars, 0, n - 1);
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reverse(chars, n, chars.length - 1);
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reverse(chars, 0, chars.length - 1);
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return new String(chars);
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}
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private void reverse(char[] c, int i, int j) {
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while(i < j) {
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char t = c[i]; c[i] = c[j]; c[j] = t;
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i++; j--;
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private void reverse(char[] chars, int i, int j) {
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while (i < j) {
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swap(chars, i++, j--);
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}
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}
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private void swap(char[] chars, int i, int j) {
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char t = chars[i];
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chars[i] = chars[j];
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chars[j] = t;
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}
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```
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# 59. 滑动窗口的最大值
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## 题目描述
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[NowCoder](https://www.nowcoder.com/practice/1624bc35a45c42c0bc17d17fa0cba788?tpId=13&tqId=11217&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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给定一个数组和滑动窗口的大小，找出所有滑动窗口里数值的最大值。例如，如果输入数组 {2, 3, 4, 2, 6, 2, 5, 1} 及滑动窗口的大小 3，那么一共存在 6 个滑动窗口，他们的最大值分别为 {4, 4, 6, 6, 6, 5}。
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## 解题思路
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@ -2414,14 +2453,13 @@ private void reverse(char[] c, int i, int j) {
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```java
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public ArrayList<Integer> maxInWindows(int[] num, int size) {
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ArrayList<Integer> ret = new ArrayList<>();
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if (size > num.length || size < 1) return ret;
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// 构建最大堆，即堆顶元素是堆的最大值。
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PriorityQueue<Integer> heap = new PriorityQueue<Integer>((o1, o2) -> o2 - o1);
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if (size > num.length || size < 1) return ret;
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for (int i = 0; i < size; i++) heap.add(num[i]);
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ret.add(heap.peek());
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for (int i = 1; i + size - 1 < num.length; i++) {
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for (int i = 1, j = i + size - 1; j < num.length; i++, j++) {
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heap.remove(num[i - 1]);
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heap.add(num[i + size - 1]);
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heap.add(num[j]);
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ret.add(heap.peek());
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}
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return ret;
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@ -2432,35 +2470,39 @@ public ArrayList<Integer> maxInWindows(int[] num, int size) {
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## 题目描述
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[Lintcode](https://www.lintcode.com/en/problem/dices-sum/)
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把 n 个骰子仍在地上，求点数和为 s 的概率。
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## 解题思路
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### 动态规划解法
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使用一个二维数组 dp 存储点数出现的次数，其中 dp[i][j] 表示前 i 个骰子产生点数 j 的次数。
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空间复杂度：O(N<sup>2</sup>)
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```java
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private static int face = 6;
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public double countProbability(int n, int s) {
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if (n < 1 || s < n) return 0.0;
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int pointNum = face * n;
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int[][] dp = new int[n][pointNum];
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for (int i = 0; i < face; i++) {
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dp[0][i] = 1;
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public List<Map.Entry<Integer, Double>> dicesSum(int n) {
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final int face = 6;
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final int pointNum = face * n;
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long[][] dp = new long[n + 1][pointNum + 1];
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for (int i = 1; i <= face; i++) {
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dp[1][i] = 1;
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}
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for (int i = 1; i < n; i++) {
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for (int j = i; j < pointNum; j++) { // 使用 i 个骰子最小点数为 i
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for (int k = 1; k <= face; k++) {
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if (j - k >= 0) {
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dp[i][j] += dp[i - 1][j - k];
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}
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for (int i = 2; i <= n; i++) {
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for (int j = i; j <= pointNum; j++) { // 使用 i 个骰子最小点数为 i
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for (int k = 1; k <= face && k <= j; k++) {
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dp[i][j] += dp[i - 1][j - k];
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}
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}
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}
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int totalNum = (int) Math.pow(6, n);
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return (double) dp[n - 1][s - 1] / totalNum;
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final double totalNum = Math.pow(6, n);
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List<Map.Entry<Integer, Double>> ret = new ArrayList<>();
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for (int i = n; i <= pointNum; i++) {
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ret.add(new AbstractMap.SimpleEntry<>(i, dp[n][i] / totalNum));
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}
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return ret;
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}
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```
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@ -2469,28 +2511,30 @@ public double countProbability(int n, int s) {
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空间复杂度：O(N)
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```java
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private static int face = 6;
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public double countProbability(int n, int s) {
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if (n < 1 || s < n) return 0.0;
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int pointNum = face * n;
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int[][] dp = new int[2][pointNum];
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for (int i = 0; i < face; i++) {
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public List<Map.Entry<Integer, Double>> dicesSum(int n) {
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final int face = 6;
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final int pointNum = face * n;
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long[][] dp = new long[2][pointNum + 1];
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for (int i = 1; i <= face; i++) {
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dp[0][i] = 1;
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}
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int flag = 1;
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for (int i = 1; i < n; i++) {
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for (int j = i; j < pointNum; j++) { // 使用 i 个骰子最小点数为 i
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for (int k = 1; k <= face; k++) {
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if (j - k >= 0) {
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dp[flag][j] += dp[1 - flag][j - k];
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}
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for (int i = 2; i <= n; i++, flag = 1 - flag) {
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for (int j = 0; j <= pointNum; j++) {
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dp[flag][j] = 0; // 旋转数组清零
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}
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for (int j = i; j <= pointNum; j++) { // 使用 i 个骰子最小点数为 i
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for (int k = 1; k <= face && k <= j; k++) {
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dp[flag][j] += dp[1 - flag][j - k];
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}
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}
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flag = 1 - flag;
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}
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int totalNum = (int) Math.pow(6, n);
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return (double) dp[flag][s - 1] / totalNum;
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final double totalNum = Math.pow(6, n);
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List<Map.Entry<Integer, Double>> ret = new ArrayList<>();
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for (int i = n; i <= pointNum; i++) {
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ret.add(new AbstractMap.SimpleEntry<>(i, dp[1 - flag][i] / totalNum));
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}
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return ret;
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}
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```
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@ -328,7 +328,7 @@ aBCd
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# 九、前后查找
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前后查找规定了匹配的内容首尾应该匹配的内容，但是又不包含首尾匹配的内容。向前查找用 **?=** 来定义，它规定了尾部匹配的内容，这个匹配的内容在 ?= 之后定义。所谓向前查找，就是规定了一个匹配的内容，然后以这个内容为尾部向前面查找需要匹配的内容。向后匹配用 ?<= 定义（注:javaScript不支持向后匹配,java对其支持也不完善）。
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前后查找规定了匹配的内容首尾应该匹配的内容，但是又不包含首尾匹配的内容。向前查找用 **?=** 来定义，它规定了尾部匹配的内容，这个匹配的内容在 ?= 之后定义。所谓向前查找，就是规定了一个匹配的内容，然后以这个内容为尾部向前面查找需要匹配的内容。向后匹配用 ?<= 定义（注: javaScript 不支持向后匹配, java 对其支持也不完善）。
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**应用**
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Loading…
Reference in New Issue
Block a user