LeetCode Search for a Range

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：进行二分搜索，当找到目标值时，在已该元素为中心进行扩展，从而确定区域范围

代码如下：

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        int middle,front,back;        int start=-1,end=-1;        vector<int> result;        front = 0,back=nums.size()-1;        middle = (front+back)/2;        while(front<=back)        {            if(nums[middle] > target)            {                back=middle-1;                middle=(front+back)/2;            }            else if(nums[middle] < target)            {                front = middle+1;                middle=(front+back)/2;            }            else            {                start=end=middle;                while(start>0 && nums[start-1] == target)                    start--;                while(end < nums.size()-1 && nums[end+1] == target)                    end++;                break;            }        }        result.push_back(start);        result.push_back(end);        return result;    }};