Search for a Range一打简单的二分题

题目描述:
Given an array of integers sorted in ascending order, 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：这是一道简单的二分题，只要考虑好边界，二分即可。
代码如下：

 vector<int> searchRange(vector<int>& nums, int target) {           int begin=0;        vector<int>temp;        if(nums.size()==0){            temp.push_back(-1);            temp.push_back(-1);            return temp;                          }        int end=nums.size()-1;        int mid=(end+begin)/2;        while(begin<end){            if(nums[mid]==target) break;            if(nums[mid]<target){                if(mid==begin){                    mid=end;                    break;                }                begin=mid;                mid=(end+begin)/2;            }            if(nums[mid]>target){                if(mid==end) {                    mid=begin;                    break;                }                end=mid;               mid=(end+begin)/2;            }        }        int a,b;        if(nums[mid]==target){            a=mid;            b=mid;            while(nums[a]==target&&a>=0){            a--;         }            while(nums[b]==target){            b++;         }        temp.push_back(a+1);        temp.push_back(b-1);    }        else{            temp.push_back(-1);            temp.push_back(-1);            return temp;        }        return temp;    }};