【LEET-CODE】34. Search for a Range【Medium】

Question:

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：

给定升序数组，找到target在其中的范围，没找到则返回【-1，-1】。

注意审题，很简单，没什么说的。

Code:

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> answer;        int x=-1,y=-1;        for(int i = 0; i < nums.size() ;i++){            if(nums[i]==target){                if(x==-1) x=i;                y = i;            }            else if(y!=-1) break;        }        answer.push_back(x);        answer.push_back(y);        return answer;    }};