Leetcode Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

二分查找

第一步找到元素的起点

第二步找到元素的终点

class Solution {
public:
    int lower_bound(int A[], int n, int target){
        int left = 0, right = n-1;
        while(left < right){
            int mid = (left+right)/2;
            if(A[mid] < target) left = mid+1;
            else right = mid;
        }
        if(A[left]!=target) return -1;
        else return left;
    }
    
    int upper_bound(int A[], int n, int target){
        int left = 0, right = n-1;
        while(left <= right){
            int mid = (left+right)/2;
            if(A[mid] > target) right = mid-1;
            else  left = mid+1;
        }
        if(A[right]!=target) return -1;
        else return right;
    }
    
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> res;
        res.push_back(lower_bound(A,n,target));
        res.push_back(upper_bound(A,n,target));
        return res;
    }
};