LeetCode 034 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].

AC代码如下：

class Solution {public://先用二分查找找出这个数，再由这个数往左、往右查找边界（非最优，最坏情况下为O(n))    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> ret;        int left=0;        int right=nums.size()-1;        int mid;        while(left<=right){            mid=(left+right)/2;            if(nums[mid]==target)                break;            else if(nums[mid]<target)                left=mid+1;            else                right=mid-1;        }                if(left>right){            ret.push_back(-1);            ret.push_back(-1);            return ret;        }else{            int i=mid,j=mid;            int length=nums.size();            for(;i>=0;i--){                if(nums[i]!=target)                    break;            }            for(;j<length;j++){                if(nums[j]!=target)                    break;            }            ret.push_back(i+1);            ret.push_back(j-1);            return ret;        }    }};