零.二分查找

二分查找

1.思路一：朴素二分

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int left=0;
        int right = nums.size()-1;

        while(left<=right)
        {
            int mid = (left+right)/2;

            //1.nums[mid] > target
            if(nums[mid]>target)
                right = mid-1;
            //2.nums[mid] < target
            else if(nums[mid]<target)
                left = mid+1;
            //3.nums[mid] == target
            else
                return mid;
        }

        return -1;
    }
};

一.在排序数组中第一个和最后一个数：

1.思路一：

特殊情况：

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int left = 0;
        int right = nums.size()-1;

        //处理边界情况
        if(nums.size() == 0)
            return {-1,-1};

        //1.确定左端点：
        while(left<right)
        {
            int mid = left + (right-left)/2;
            if(nums[mid] < target) left = mid+1;
            else if(nums[mid] >= target) right = mid;
        }
        if(nums[left]!=target) return {-1,-1};

        int begin = left;
        //2.确定右端点：
        left=0,right=nums.size()-1;
        while(left<right)
        {
            int mid = left+(right-mid+1)/2;
            if(nums[mid] <= target) left = mid;
            else if(nums[mid] > target) right = mid-1;
        }

        int end = left;
        return {begin,end};
    }
};

GIF题目解析

找左端点：

找右端点：

二.算法X的平方根：

X的平方根

1.思路一：暴力+哈希

2.思路二：二分区间

class Solution {
public:
    int mySqrt(int x) {
        if(x==0)
            return 0;

        long long  left = 1;
        long long  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;
    }
};

GIF题目解析

三.搜索插入位置：

搜索插入位置

1.思路一：

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

        if(nums.size() == 1)
        {
            if(target > nums[0])
                return 1;
            if(target <= nums[0])
                return 0;
        }
        while(left < right)
        {
            int mid = left + (right - left)/2;
            if(nums[mid] < target) left = mid+1;
            else right = mid;
        }

        if(nums[right] < target) return right + 1;
        return right;
    }
};

GIF题目解析

四：山脉数组的峰顶索引：

山脉数组的峰顶索引

1.思路一：

class Solution {
public:
    int peakIndexInMountainArray(vector<int>& arr) {
        int left=0,right=arr.size()-1;

        while(left<right)
        {
            int mid = left + (right-left +1)/2;
            if(arr[mid-1] < arr[mid]) left=mid;
            else right = mid -1;
        }
        return left;
    }
};

GIF题目解析

五：寻找峰值：

寻找峰值

1.思路一：

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

GIF题目解析

六：寻找旋转排序数组中的最小值

寻找旋转排序数组中的最小值

1.思路一：

class Solution {
public:
    int findMin(vector<int>& nums) {

        int left = 0, right = nums.size() - 1;
        int n = nums[right];

        while (left < right)
        {
            int mid = left + (right - left) / 2;
            if (nums[mid] > n) left = mid + 1;
            else right = mid;
        }

        return nums[right];
    }
};

GIF题目解析

七：0~~n-1中缺少的数字(点名)：

点名

1.思路一：

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(mid == records[mid]) left = mid +1;
            else right = mid;
        }
        if(records[left] == left)
            return left+1;
            
        return left;
    }
};

GIF题目解析