Search for a Range <leetcode>

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].

 

算法：该算法来源于网络，用二分查找最左最右的位置，代码如下：

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        int l=findPos(A,0,n-1,target,true);
        int r=findPos(A,0,n-1,target,false);
        vector<int> result;
        result.push_back(l);
        result.push_back(r);
        return result;
    }
    int findPos(int a[],int beg,int end,int key,bool findLeft)
    {
        if(beg>end)   return -1;
        int mid=(beg+end)/2;
        if(a[mid]==key)
        {
            int pos=findLeft?findPos(a,beg,mid-1,key,findLeft):findPos(a,mid+1,end,key,findLeft);
            return pos==-1?mid:pos;
        }
        else if(a[mid]<key)
        {
            findPos(a,mid+1,end,key,findLeft);
        }
        else if(a[mid]>key)
        {
            findPos(a,beg,mid-1,key,findLeft);
        }
    }
};