二分查找

Author: yangcongrui2022

data_structure/二分查找.md

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+## List-binarysearch
+### 二分查找
+#### 解题步骤
+1. 确定二分左右边界
+2. 设计一个性质使得可以将区间划分成两段（要求的点在其中一段的端点）
+3. 区间更新
+#### 二分查找模板
+```
+bool check(int x) // 检查x是否满足某种性质
+int bsearch_1(int l, int r){
+    while (l < r){
+        int mid = l + r >> 1;
+        if (check(mid)) r = mid;
+        else l = mid + 1;
+    }
+    return l;
+}
+
+int bsearch_2(int l, int r){
+    while (l < r){
+        int mid = l + r + 1>> 1;
+        if (check(mid)) l = mid;
+        else r = mid - 1;
+    }
+    return l;
+}
+```
+#### leetcode.704  
+- 链接<https://leetcode.cn/problems/binary-search/>  
+- leetcode解题代码
+```
+class Solution {
+public:
+    int search(vector<int>& nums, int target) {
+        int l = 0, r = nums.size() - 1;
+        while (l < r){
+            int mid = (l + r) / 2;
+            if (nums[mid] >= target) r = mid;
+            else l = mid + 1;
+        }
+        if (nums[l] == target) return l;
+        return -1;
+    }
+};
+```
+#### leetcode.278  
+- 链接<https://leetcode.cn/problems/first-bad-version/>  
+- leetcode解题代码
+```
+// The API isBadVersion is defined for you.
+// bool isBadVersion(int version);
+
+class Solution {
+public:
+    int firstBadVersion(int n) {
+        long l = 1, r = n;
+        while (l < r){
+            int mid = (l + r) / 2;
+            if (isBadVersion(mid)) r = mid;
+            else l = mid + 1;
+        }
+        return l;// 返回的是第一个错误的版本
+    }
+};
+```
+#### leetcode.35  
+- 链接<https://leetcode.cn/problems/search-insert-position/>  
+- leetcode解题代码
+```
+class Solution {
+public:
+    int searchInsert(vector<int>& nums, int target) {
+        int n = nums.size();
+        if (target > nums[n - 1]) return n;
+        int l = 0, r = n - 1;
+        while (l < r){
+            int mid = (l + r) / 2;
+            if (nums[mid] >= target) r = mid;
+            else l = mid + 1;
+        }
+        return l;
+    }
+};
+```
+#### leetcode.69  
+- 链接<https://leetcode.cn/problems/sqrtx/submissions/>  
+- leetcode解题代码
+```
+class Solution {
+public:
+    int mySqrt(int x) {
+        if (x <= 1) return x;
+        long l = 1, r = x;
+        while (l < r){
+            int mid = l + r + 1 >> 1;
+            if (mid <= x / mid) l = mid;// 下取整，取第一个小于等于根号下x的值
+            else r = mid - 1;
+        }
+
+        return l;
+    }
+};
+```
+#### leetcode.367  
+- 链接<https://leetcode.cn/problems/valid-perfect-square/>  
+- 注意：与上一题区别在于只是判断是否存在完全平方数，不需要上取整或者下取整求完全平方数，所以使用模板一或者模板二均可  
+- leetcode解题代码
+```
+class Solution {
+public:
+    bool isPerfectSquare(int num) {
+        long l = 1, r = num;
+        while (l < r){
+            int mid = l + r + 1 >> 1;
+            if (mid <= num / mid) l = mid;
+            else r = mid - 1;
+        }
+        return r * r == num;
+    }
+};
+```
+#### leetcode.34
+- 链接<https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/>  
+- leetcode解题代码
+```
+class Solution {
+public:
+    vector<int> searchRange(vector<int>& nums, int target) {
+        if (nums.empty()) return {-1, -1};
+        int l = 0, r = nums.size() - 1;
+        while (l < r){
+            int mid = l + r >> 1;
+            // 找到>=target的第一个数
+            if (nums[mid] >= target) r = mid;
+            else l = mid + 1;
+        }
+
+        if (nums[l] != target) return {-1, -1};
+        int L = l;
+
+        l = 0, r = nums.size() - 1;
+        while (l < r){
+            int mid = l + r + 1 >> 1;
+            // 找到<=target的第一个数
+            if (nums[mid] <= target) l = mid;
+            else r = mid - 1;
+        }
+        return {L, r};
+    }
+};
+```
+#### leetcode.74
+- 链接<https://leetcode.cn/problems/search-a-2d-matrix/>  
+- 解题思路：将二维矩阵展成一维数组，二分
+- 注意：矩阵下标（i，j）转化为一维数组下标
+- leetcode解题代码
+```
+class Solution {
+public:
+    bool searchMatrix(vector<vector<int>>& matrix, int target) {
+        if (matrix.empty()) return false;
+        int n = matrix.size(), m = matrix[0].size();
+        int l = 0, r = n * m - 1;
+        while (l < r){
+            int mid = l + r >> 1;
+            // 将矩阵下标转换为数组下标
+            if (matrix[mid / m][mid % m] >= target) r = mid;
+            else l = mid + 1;
+        }
+
+        if (matrix[l / m][l % m] == target) return true;
+        return false;
+    }
+};
+```
+#### leetcode.153
+- 链接<https://leetcode.cn/problems/find-minimum-in-rotated-sorted-array/>  
+- 解题思路：找到数组的二段性（前段的值均大于数组的最后一位，后段的值均小于等于数组的最后一位，并且最小的点在后段的端点处）
+- leetcode解题代码
+```
+class Solution {
+public:
+    int findMin(vector<int>& nums) {
+        int l = 0, r = nums.size();
+        while (l < r){
+            int mid = l + r >> 1;
+            if (nums[mid] <= nums.back()) r = mid;
+            else l = mid + 1;
+        }
+        return nums[l];
+    }
+};
+```
+#### leetcode.33
+- 链接<https://leetcode.cn/problems/search-in-rotated-sorted-array/>  
+- 解题思路：由于没有二段性不能直接二分，先找到数组的最小值（参考上一题），找到目标值所在区间，在对应区间内二分找目标值
+- leetcode解题代码
+```
+class Solution {
+public:
+    int search(vector<int>& nums, int target) {
+        if (nums.empty()) return -1;
+        // 找到最小值
+        int l = 0, r = nums.size() - 1;
+        while (l < r){
+            int mid = (l + r) / 2;
+            if (nums[mid] <= nums.back()) r = mid;
+            else l = mid + 1;
+        }
+        // 判断目标值所在区间
+        if (target <= nums.back()) r = nums.size() - 1;
+        else l = 0, r --;
+        // 在所在区间找目标值
+        while (l < r){
+            int mid = (l + r) / 2;
+            if (nums[mid] >= target) r = mid;
+            else l = mid + 1;
+        }
+        if (nums[l] == target) return l;
+        return -1;
+    }
+};
+```
+***
+下列题目没有二段性，但仍然可以通过二分每次缩小一般的搜索区间找到答案
+#### leetcode.162
+- 链接<https://leetcode.cn/problems/find-peak-element/>  
+- 解题思路：如果二分的中点的值比中点右边的值小，那么右边一定存在峰值，同理，如果二分的中点的值比中点右边的值大，那么左边一定存在峰值
+- leetcode解题代码
+```
+class Solution {
+public:
+    int findPeakElement(vector<int>& nums) {
+        int l = 0, r = nums.size() - 1;
+        while (l < r){
+            int mid = l + r >> 1;
+            // 说明左边一定有峰值
+            if (nums[mid] > nums[mid + 1]) r = mid;
+            else l = mid + 1;
+        }
+        return l;
+    }
+};
+
+class Solution {
+public:
+    int findPeakElement(vector<int>& nums) {
+        int l = 0, r = nums.size() - 1;
+        while (l < r){
+            int mid = l + r >> 1;
+            // 说明右边一定有峰值
+            if (nums[mid] < nums[mid + 1]) l = mid + 1;
+            else r = mid;
+        }
+        return l;
+    }
+};
+```
+#### leetcode.287
+- 链接<https://leetcode.cn/problems/find-the-duplicate-number/>  
+- 解题思路：抽屉原理，n+1个苹果放进n个抽屉中必然有一个抽屉放了两个苹果，抽屉为数据范围n，苹果为数字总数n+1，对数据范围二分，记录二分中点左侧的数字总数，和左边数据范围对比，如果数字总数大于数据范围，说明重复数在左侧，右侧同理
+- leetcode解题代码
+```
+class Solution {
+public:
+    int findDuplicate(vector<int>& nums) {
+        int l = 1, r = nums.size();
+        while (l < r){
+            int mid = l + r >> 1;
+
+            int cnt = 0;
+            for (auto c: nums){
+                if (c >= l && c <= mid)
+                    cnt ++;
+            }
+            if (cnt > mid - l + 1) r = mid;
+            else l = mid + 1;
+        }
+        return l;
+    }
+};
+```
+解题参考：<https://www.acwing.com/>