Search for a Range

题目：
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].
思路：
1、利用二分法找到为target的值；
2、再用二分法找到最小的begin和最大的end；
代码：

class Solution {public:    vector<int> searchRange(int A[], int n, int target) {        begin=-1;        end=-1;        vector<int> res;        find(A,0,n-1,target);        res.push_back(begin);        res.push_back(end);        return res;    }private:   int begin=-1;   int end=-1;   void find(int A[],int left,int right,int target)   {       if(left>right) return;       int mid=(left+right)>>1;       if(A[mid]==target)       {           if(mid<begin || begin==-1) begin=mid;           if(mid>end) end=mid;           find(A,left,mid-1,target);           find(A,mid+1,right,target);       }       else if(A[mid]>target) find(A,left,mid-1,target);       else if(A[mid]<target) find(A,mid+1,right,target);   }};