题解 | #二分查找-I#（3种方法）

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int len = nums.size();
        int left=0, right=len-1;
        while(left <= right){
            mid = left+(right-left >> 1);  // +-大于位移运算的优先级 左移*2
            if(nums[mid] == target) return mid;
            if(nums[mid] < target) left = mid+1;
            if(nums[mid] > target) right = mid-1;
        }
        return -1;
    }
};

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int right = nums.size();
        if(right==0) return -1;
        int left = 0;
        while(left < right){
            int mid = left + right >>1;
            nums[mid]<target?left=mid+1:right=mid;  // lleft为找到的大于等于target的下标
        }
        return nums[left]==target?left:-1;
    }
};

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int right = nums.size();
        if(right==0) return -1;
        int ans = upper_bound(nums.begin(), nums.end(), target) - nums.begin() - 1;
        return nums[ans]==target?ans:-1;
    }
};