2019-04-25 18:24:51 +08:00
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<!-- GFM-TOC -->
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* [40. 最小的 K 个数](#40-最小的-k-个数)
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* [41.1 数据流中的中位数](#411-数据流中的中位数)
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* [41.2 字符流中第一个不重复的字符](#412-字符流中第一个不重复的字符)
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* [42. 连续子数组的最大和](#42-连续子数组的最大和)
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* [43. 从 1 到 n 整数中 1 出现的次数](#43-从-1-到-n-整数中-1-出现的次数)
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* [44. 数字序列中的某一位数字](#44-数字序列中的某一位数字)
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* [45. 把数组排成最小的数](#45-把数组排成最小的数)
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* [46. 把数字翻译成字符串](#46-把数字翻译成字符串)
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* [47. 礼物的最大价值](#47-礼物的最大价值)
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* [48. 最长不含重复字符的子字符串](#48-最长不含重复字符的子字符串)
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* [49. 丑数](#49-丑数)
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<!-- GFM-TOC -->
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# 40. 最小的 K 个数
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[NowCoder](https://www.nowcoder.com/practice/6a296eb82cf844ca8539b57c23e6e9bf?tpId=13&tqId=11182&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 解题思路
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### 快速选择
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- 复杂度：O(N) + O(1)
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- 只有当允许修改数组元素时才可以使用
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快速排序的 partition() 方法，会返回一个整数 j 使得 a[l..j-1] 小于等于 a[j]，且 a[j+1..h] 大于等于 a[j]，此时 a[j] 就是数组的第 j 大元素。可以利用这个特性找出数组的第 K 个元素，这种找第 K 个元素的算法称为快速选择算法。
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```java
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public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k) {
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ArrayList<Integer> ret = new ArrayList<>();
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if (k > nums.length || k <= 0)
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return ret;
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findKthSmallest(nums, k - 1);
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/* findKthSmallest 会改变数组，使得前 k 个数都是最小的 k 个数 */
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for (int i = 0; i < k; i++)
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ret.add(nums[i]);
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return ret;
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}
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public void findKthSmallest(int[] nums, int k) {
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int l = 0, h = nums.length - 1;
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while (l < h) {
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int j = partition(nums, l, h);
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if (j == k)
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break;
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if (j > k)
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h = j - 1;
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else
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l = j + 1;
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}
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}
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private int partition(int[] nums, int l, int h) {
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int p = nums[l]; /* 切分元素 */
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int i = l, j = h + 1;
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while (true) {
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while (i != h && nums[++i] < p) ;
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while (j != l && nums[--j] > p) ;
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if (i >= j)
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break;
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swap(nums, i, j);
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}
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swap(nums, l, j);
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return j;
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}
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private void swap(int[] nums, int i, int j) {
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int t = nums[i];
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nums[i] = nums[j];
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nums[j] = t;
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}
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```
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### 大小为 K 的最小堆
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- 复杂度：O(NlogK) + O(K)
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- 特别适合处理海量数据
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应该使用大顶堆来维护最小堆，而不能直接创建一个小顶堆并设置一个大小，企图让小顶堆中的元素都是最小元素。
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维护一个大小为 K 的最小堆过程如下：在添加一个元素之后，如果大顶堆的大小大于 K，那么需要将大顶堆的堆顶元素去除。
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```java
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public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k) {
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if (k > nums.length || k <= 0)
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return new ArrayList<>();
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PriorityQueue<Integer> maxHeap = new PriorityQueue<>((o1, o2) -> o2 - o1);
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for (int num : nums) {
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maxHeap.add(num);
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if (maxHeap.size() > k)
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maxHeap.poll();
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}
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return new ArrayList<>(maxHeap);
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}
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```
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# 41.1 数据流中的中位数
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[NowCoder](https://www.nowcoder.com/practice/9be0172896bd43948f8a32fb954e1be1?tpId=13&tqId=11216&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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如何得到一个数据流中的中位数？如果从数据流中读出奇数个数值，那么中位数就是所有数值排序之后位于中间的数值。如果从数据流中读出偶数个数值，那么中位数就是所有数值排序之后中间两个数的平均值。
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## 解题思路
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```java
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/* 大顶堆，存储左半边元素 */
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private PriorityQueue<Integer> left = new PriorityQueue<>((o1, o2) -> o2 - o1);
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/* 小顶堆，存储右半边元素，并且右半边元素都大于左半边 */
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private PriorityQueue<Integer> right = new PriorityQueue<>();
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/* 当前数据流读入的元素个数 */
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private int N = 0;
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public void Insert(Integer val) {
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/* 插入要保证两个堆存于平衡状态 */
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if (N % 2 == 0) {
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/* N 为偶数的情况下插入到右半边。
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* 因为右半边元素都要大于左半边，但是新插入的元素不一定比左半边元素来的大，
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* 因此需要先将元素插入左半边，然后利用左半边为大顶堆的特点，取出堆顶元素即为最大元素，此时插入右半边 */
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left.add(val);
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right.add(left.poll());
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} else {
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right.add(val);
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left.add(right.poll());
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}
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N++;
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}
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public Double GetMedian() {
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if (N % 2 == 0)
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return (left.peek() + right.peek()) / 2.0;
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else
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return (double) right.peek();
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}
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```
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# 41.2 字符流中第一个不重复的字符
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[NowCoder](https://www.nowcoder.com/practice/00de97733b8e4f97a3fb5c680ee10720?tpId=13&tqId=11207&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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请实现一个函数用来找出字符流中第一个只出现一次的字符。例如，当从字符流中只读出前两个字符 "go" 时，第一个只出现一次的字符是 "g"。当从该字符流中读出前六个字符“google" 时，第一个只出现一次的字符是 "l"。
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## 解题思路
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```java
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private int[] cnts = new int[256];
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private Queue<Character> queue = new LinkedList<>();
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public void Insert(char ch) {
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cnts[ch]++;
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queue.add(ch);
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while (!queue.isEmpty() && cnts[queue.peek()] > 1)
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queue.poll();
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}
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public char FirstAppearingOnce() {
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return queue.isEmpty() ? '#' : queue.peek();
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}
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```
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# 42. 连续子数组的最大和
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[NowCoder](https://www.nowcoder.com/practice/459bd355da1549fa8a49e350bf3df484?tpId=13&tqId=11183&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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{6, -3, -2, 7, -15, 1, 2, 2}，连续子数组的最大和为 8（从第 0 个开始，到第 3 个为止）。
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## 解题思路
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```java
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public int FindGreatestSumOfSubArray(int[] nums) {
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if (nums == null || nums.length == 0)
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return 0;
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int greatestSum = Integer.MIN_VALUE;
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int sum = 0;
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for (int val : nums) {
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sum = sum <= 0 ? val : sum + val;
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greatestSum = Math.max(greatestSum, sum);
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}
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return greatestSum;
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}
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```
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# 43. 从 1 到 n 整数中 1 出现的次数
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[NowCoder](https://www.nowcoder.com/practice/bd7f978302044eee894445e244c7eee6?tpId=13&tqId=11184&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 解题思路
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```java
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public int NumberOf1Between1AndN_Solution(int n) {
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int cnt = 0;
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for (int m = 1; m <= n; m *= 10) {
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int a = n / m, b = n % m;
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cnt += (a + 8) / 10 * m + (a % 10 == 1 ? b + 1 : 0);
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}
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return cnt;
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}
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```
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> [Leetcode : 233. Number of Digit One](https://leetcode.com/problems/number-of-digit-one/discuss/64381/4+-lines-O(log-n)-C++JavaPython)
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# 44. 数字序列中的某一位数字
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## 题目描述
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数字以 0123456789101112131415... 的格式序列化到一个字符串中，求这个字符串的第 index 位。
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## 解题思路
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```java
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public int getDigitAtIndex(int index) {
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if (index < 0)
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return -1;
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int place = 1; // 1 表示个位，2 表示 十位...
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while (true) {
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int amount = getAmountOfPlace(place);
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int totalAmount = amount * place;
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if (index < totalAmount)
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return getDigitAtIndex(index, place);
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index -= totalAmount;
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place++;
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}
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}
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/**
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* place 位数的数字组成的字符串长度
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* 10, 90, 900, ...
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*/
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private int getAmountOfPlace(int place) {
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if (place == 1)
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return 10;
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return (int) Math.pow(10, place - 1) * 9;
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}
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/**
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* place 位数的起始数字
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* 0, 10, 100, ...
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*/
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private int getBeginNumberOfPlace(int place) {
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if (place == 1)
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return 0;
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return (int) Math.pow(10, place - 1);
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}
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/**
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* 在 place 位数组成的字符串中，第 index 个数
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*/
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private int getDigitAtIndex(int index, int place) {
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int beginNumber = getBeginNumberOfPlace(place);
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int shiftNumber = index / place;
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String number = (beginNumber + shiftNumber) + "";
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int count = index % place;
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return number.charAt(count) - '0';
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}
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```
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# 45. 把数组排成最小的数
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[NowCoder](https://www.nowcoder.com/practice/8fecd3f8ba334add803bf2a06af1b993?tpId=13&tqId=11185&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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输入一个正整数数组，把数组里所有数字拼接起来排成一个数，打印能拼接出的所有数字中最小的一个。例如输入数组 {3，32，321}，则打印出这三个数字能排成的最小数字为 321323。
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## 解题思路
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可以看成是一个排序问题，在比较两个字符串 S1 和 S2 的大小时，应该比较的是 S1+S2 和 S2+S1 的大小，如果 S1+S2 < S2+S1，那么应该把 S1 排在前面，否则应该把 S2 排在前面。
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```java
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public String PrintMinNumber(int[] numbers) {
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if (numbers == null || numbers.length == 0)
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return "";
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int n = numbers.length;
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String[] nums = new String[n];
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for (int i = 0; i < n; i++)
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nums[i] = numbers[i] + "";
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Arrays.sort(nums, (s1, s2) -> (s1 + s2).compareTo(s2 + s1));
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String ret = "";
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for (String str : nums)
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ret += str;
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return ret;
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}
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```
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# 46. 把数字翻译成字符串
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[Leetcode](https://leetcode.com/problems/decode-ways/description/)
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## 题目描述
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给定一个数字，按照如下规则翻译成字符串：1 翻译成“a”，2 翻译成“b”... 26 翻译成“z”。一个数字有多种翻译可能，例如 12258 一共有 5 种，分别是 abbeh，lbeh，aveh，abyh，lyh。实现一个函数，用来计算一个数字有多少种不同的翻译方法。
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## 解题思路
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```java
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public int numDecodings(String s) {
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if (s == null || s.length() == 0)
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return 0;
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int n = s.length();
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int[] dp = new int[n + 1];
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dp[0] = 1;
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dp[1] = s.charAt(0) == '0' ? 0 : 1;
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for (int i = 2; i <= n; i++) {
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int one = Integer.valueOf(s.substring(i - 1, i));
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if (one != 0)
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dp[i] += dp[i - 1];
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if (s.charAt(i - 2) == '0')
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continue;
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int two = Integer.valueOf(s.substring(i - 2, i));
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if (two <= 26)
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dp[i] += dp[i - 2];
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}
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return dp[n];
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}
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```
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# 47. 礼物的最大价值
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[NowCoder](https://www.nowcoder.com/questionTerminal/72a99e28381a407991f2c96d8cb238ab)
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## 题目描述
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在一个 m\*n 的棋盘的每一个格都放有一个礼物，每个礼物都有一定价值（大于 0）。从左上角开始拿礼物，每次向右或向下移动一格，直到右下角结束。给定一个棋盘，求拿到礼物的最大价值。例如，对于如下棋盘
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```
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1 10 3 8
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12 2 9 6
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5 7 4 11
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3 7 16 5
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```
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礼物的最大价值为 1+12+5+7+7+16+5=53。
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## 解题思路
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应该用动态规划求解，而不是深度优先搜索，深度优先搜索过于复杂，不是最优解。
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```java
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public int getMost(int[][] values) {
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if (values == null || values.length == 0 || values[0].length == 0)
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return 0;
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int n = values[0].length;
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int[] dp = new int[n];
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for (int[] value : values) {
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dp[0] += value[0];
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for (int i = 1; i < n; i++)
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dp[i] = Math.max(dp[i], dp[i - 1]) + value[i];
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}
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return dp[n - 1];
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}
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```
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# 48. 最长不含重复字符的子字符串
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## 题目描述
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输入一个字符串（只包含 a\~z 的字符），求其最长不含重复字符的子字符串的长度。例如对于 arabcacfr，最长不含重复字符的子字符串为 acfr，长度为 4。
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## 解题思路
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```java
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public int longestSubStringWithoutDuplication(String str) {
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int curLen = 0;
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int maxLen = 0;
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int[] preIndexs = new int[26];
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Arrays.fill(preIndexs, -1);
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for (int curI = 0; curI < str.length(); curI++) {
|
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int c = str.charAt(curI) - 'a';
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int preI = preIndexs[c];
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|
if (preI == -1 || curI - preI > curLen) {
|
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curLen++;
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|
} else {
|
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maxLen = Math.max(maxLen, curLen);
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|
curLen = curI - preI;
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}
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|
|
preIndexs[c] = curI;
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}
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maxLen = Math.max(maxLen, curLen);
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|
return maxLen;
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}
|
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```
|
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# 49. 丑数
|
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|
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|
|
[NowCoder](https://www.nowcoder.com/practice/6aa9e04fc3794f68acf8778237ba065b?tpId=13&tqId=11186&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
|
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## 题目描述
|
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|
把只包含因子 2、3 和 5 的数称作丑数（Ugly Number）。例如 6、8 都是丑数，但 14 不是，因为它包含因子 7。习惯上我们把 1 当做是第一个丑数。求按从小到大的顺序的第 N 个丑数。
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|
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## 解题思路
|
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|
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|
|
```java
|
|
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|
|
public int GetUglyNumber_Solution(int N) {
|
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|
|
if (N <= 6)
|
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|
|
return N;
|
|
|
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|
|
int i2 = 0, i3 = 0, i5 = 0;
|
|
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|
|
int[] dp = new int[N];
|
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|
|
dp[0] = 1;
|
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|
|
for (int i = 1; i < N; i++) {
|
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|
|
int next2 = dp[i2] * 2, next3 = dp[i3] * 3, next5 = dp[i5] * 5;
|
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|
|
dp[i] = Math.min(next2, Math.min(next3, next5));
|
|
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|
|
if (dp[i] == next2)
|
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|
|
i2++;
|
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|
|
if (dp[i] == next3)
|
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|
|
i3++;
|
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|
|
if (dp[i] == next5)
|
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|
|
i5++;
|
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|
|
}
|
|
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|
|
return dp[N - 1];
|
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|
|
}
|
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|
|
```
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2019-05-15 12:19:11 +08:00
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</br><div align="center">😢 ☹️🙁😐🙂😄😂🤣 </br></br> 更多精彩内容将发布在公众号 **CyC2018**，公众号提供了该项目的离线阅读版本，后台回复"下载" 即可领取。也提供了一份技术面试复习思维导图，不仅系统整理了面试知识点，而且标注了各个知识点的重要程度，从而帮你理清多而杂的面试知识点，后台回复"资料" 即可领取。我基本是按照这个思维导图来进行复习的，对我拿到了 BAT 头条等 Offer 起到很大的帮助。你们完全可以和我一样根据思维导图上列的知识点来进行复习，就不用看很多不重要的内容，也可以知道哪些内容很重要从而多安排一些复习时间。</div></br>
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2019-04-25 18:24:51 +08:00
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<div align="center"><img width="180px" src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/%E5%85%AC%E4%BC%97%E5%8F%B7.jpg"></img></div>
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