Search for a Range 寻找一个区间 二分查找左边展开

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]​​.

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
    //二分查找 找到后左右展开
    int i,left,right,mid;
    vector<int> res;
    left=0;right=n-1;
    while(left<=right)
    {
        mid=left+(right-left)/2;
        if(A[mid]<target)
           left=mid+1;
        else if(A[mid]>target)
           right=mid-1;
        else
        {
            for(i=mid;i>0;i--)
            {
                if(A[i-1]!=A[i])
                   break;
            }
            res.push_back(i);
            for(i=mid;i<n-1;i++)
            {
                if(A[i+1]!=A[i])
                   break;
            }
            res.push_back(i);
            return res;
        }
    }
    res.push_back(-1);
    res.push_back(-1);
    return res;
    }
};