【leetcode】二分查找专题

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