【C++算法】二分查找

 二分查找

• 题目链接

二分查找https://leetcode.cn/problems/binary-search/

• 算法原理

• 代码步骤

• 代码展示

class Solution {
public:int search(vector<int>& nums, int target) {int left = 0, right = nums.size() - 1;while(left <= right){// 防止溢出int mid = left + (right - left + 1) / 2;if(nums[mid] < target) {left = mid + 1;}else if(nums[mid] > target){right = mid - 1;}else {return mid;}}return -1;}
};

在排序数组中查找元素的第一个和最后一个位置

• 题目链接

在排序数组中查找元素的第一个和最后一个位置https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/

• 算法原理

• 代码步骤

• 代码展示

class Solution {
public:vector<int> searchRange(vector<int>& nums, int target) {if(nums.size() == 0){return {-1, -1};}// 设置左端点和右端点int begin = -1, end = -1;int left = 0, right = nums.size() - 1;while(left < right){int mid = left + (right - left) / 2;if(nums[mid] < target){left = mid + 1;}else {right = mid;}}if(nums[left] != target) return {-1, -1};begin = left;left = 0, right = nums.size() - 1;while(left < right){int mid = left + (right - left + 1) / 2;if(nums[mid] <= target){left = mid;}else {right = mid - 1;}}if(nums[right] != target) return {-1, -1};end = right;return {begin, end};}
};

x的平方根

• 题目链接

x的平方根https://leetcode.cn/problems/sqrtx/

• 算法原理

• 代码展示

class Solution {
public:int mySqrt(int x) {if(x == 0) return 0;int left = 1, right = x;while(left < right){long long mid = left + (right - left + 1) / 2;if(mid * mid <= x){left = mid;}else {right = mid - 1;}}return left;}
};

搜索插入位置

• 题目链接

搜索插入位置https://leetcode.cn/problems/search-insert-position/description/

• 算法原理

• 代码展示

class Solution {
public:int searchInsert(vector<int>& nums, int target) {int left = 0, right = nums.size() - 1;while(left < right){int mid = left + (right - left) / 2;if(nums[mid] < target){left = mid + 1;}else{right = mid;}}if(nums[left] < target) return left + 1;return left;}
};

山脉数组的峰顶索引

• 题目链接

山脉数组的峰顶索引https://leetcode.cn/problems/peak-index-in-a-mountain-array/description/

• 算法原理

• 代码展示

class Solution {
public:int peakIndexInMountainArray(vector<int>& arr) {// 最左侧和最右侧元素不可能使山顶int left = 1, right = arr.size() - 2;while(left < right){int mid = left + (right - left + 1) / 2;if(arr[mid] > arr[mid - 1]){left = mid;}else{right = mid - 1;}}return left;}
};

寻找峰值

• 题目链接

寻找峰值https://leetcode.cn/problems/find-peak-element/

• 算法原理

• 代码展示

class Solution {
public:int findPeakElement(vector<int>& nums) {int left = 0, right = nums.size() - 1;while(left < right){int mid = left + (right - left) / 2;if(nums[mid] < nums[mid+1]){left = mid + 1;}else{right = mid;}}return left;}
};

寻找旋转排序数组中的最小值

• 题目链接

寻找旋转排序数组中的最小值https://leetcode.cn/problems/find-minimum-in-rotated-sorted-array/description/

• 算法原理

• 代码展示

class Solution {
public:int findMin(vector<int>& nums) {int left = 0, right = nums.size() - 1;int n = nums.size();while(left < right){int mid = left + (right - left) / 2;if(nums[mid] > nums[n - 1]){left = mid + 1;}else{right = mid;}}return nums[left];}
};

点名

• 题目链接

点名https://leetcode.cn/problems/que-shi-de-shu-zi-lcof/description/

• 算法原理

• 代码展示

class Solution {
public:int takeAttendance(vector<int>& records) {int left = 0, right = records.size() - 1;while(left < right){int mid = left + (right - left) / 2;if(records[mid] == mid){left = mid + 1;}else{right = mid;}}// if(left == records.size() - 1 && records[left] == left) return left + 1;// else return left;return left == records[left] ? left + 1 : left;}
};