leetcode 34 Search for a Range(二分法)

Search for a Range

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].

题解：很简单，lower_bound()和upper_bound()两个函数用一下就好。

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int>ans;
        int n=nums.size();
        int a=lower_bound(nums.begin(),nums.end(),target)-nums.begin();
        if((a==n)||(nums[a]!=target)){
            ans.push_back(-1);
            ans.push_back(-1);
            return ans;
        }
        else{
            int b=upper_bound(nums.begin(),nums.end(),target)-nums.begin();
            ans.push_back(a);
            ans.push_back(b-1);
            return ans;
        }
    }
};