leetcode 34. Search for a Range

1.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].

2.思路 

二分查找，然后向两边延伸。。。

class Solution {public:    int bSearch(vector<int> &nums,int target){        int left = 0,right = nums.size()-1,mid = 0;        while(left <= right){            mid = (right - left)/2 + left;            if(nums[mid] == target)                return mid;            else if(nums[mid] > target)                right = mid - 1;            else                left = mid + 1;        }        return -1;    }    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> ans(2,-1);        int index = bSearch(nums,target);        if(index == -1) return ans;        else{            int left = index,right = index;            while(left >=0 && nums[left] == target)                left--;            while(right < nums.size() && nums[right] == target)                right++;            ans[0] = left+1;            ans[1] = right-1;            return ans;        }    }};