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;
        }
    }
};