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：二分法。这道题还是二分查找法，套用经典的模版，注意两次查找，第一次更新high，找到最左边的索引，第二次更新low，找到最右边的索引。

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