126企业邮箱入口_汕头人才引进_seo关键字优化价格_推广普通话的意义

1.二分查找

1.题目

题目链接

2.解题思路

3. 解题代码

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

2.在排序数组中查找元素的第一个和最后一个位置

1.题目

题目链接

2.算法原理

3. 代码

class Solution {
public:vector<int> searchRange(vector<int>& nums, int target) {int n = nums.size(), left = 0, right = n - 1;if(n == 0) return {-1, -1};// 查找左端点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};int left1 = left;right = n - 1;//找右端点while(left < right){int mid = left + (right - left + 1) / 2;if(nums[mid] <= target) left = mid;else right = mid - 1;}return {left1, left};}
};

3.x的平方根

1.题目

题目链接

2.代码

class Solution {
public:int mySqrt(int x) {long long left = 0, 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;}
};

4.搜索插入位置

1.题目

题目链接

2.解题思路

3.解题代码

class Solution {
public:int searchInsert(vector<int>& nums, int target) {int n = nums.size(), left = 0, right = n - 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; else return left + 1;}
};

5.山脉数组的索引

1.题目

题目链接

2.解题思路

3. 代码

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

6.寻找峰值

1.题目

题目链接

2.解题思路

3.代码

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

7. 寻找旋转排序数组中的最小值

7.1 题目

题目链接

7.2 解题思路

7.3 代码

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

// 原来看题解的代码
class Solution {
public:int findMin(vector<int>& nums) {int left = -1;int right = nums.size() - 1;while(left + 1 < right){int mid = left + (right - left) / 2;if (nums[mid] < nums.back()){right = mid;}else{left= mid;}}return nums[right];}
};

8.0~n-1中缺失的数字

1.题目

（剑指offer的题目，所以不能免费做题）

2.思路

3.代码