【一天一道LeetCode】#34. Search for a Range

一天一道LeetCode系列

（一）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].

（二）解题

/*首先二分法查找目标值然后沿着目标值左右延伸，依次找到相同值得左右边界*/class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        int i = 0;        int j = nums.size()-1;        int idx = -1;        vector<int> ret;        while(i <= j)        {            int mid = (i+j)/2;            if(nums[mid] == target)            {                idx = mid;                break;            }            else if(nums[mid]>target)            {                j = mid-1;            }            else if(nums[mid]<target)            {                i = mid+1;            }        }        if(idx!=-1){            int k = idx;            while(k>=0 && nums[k] == target) k--;            int m = idx;            while(m<nums.size()&&nums[m] == target) m++;            ret.push_back(k+1);            ret.push_back(m-1);        }        else{            ret.push_back(-1);            ret.push_back(-1);        }        return ret;    }};