LeetCode Array Partition I

LeetCode Array Partition I

Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.

Example 1:

Input: [1,4,3,2]

Output: 4
Explanation: n is 2, and the maximum sum of pairs is 4.

Note:

  1. n is a positive integer, which is in the range of [1, 10000].
  2. All the integers in the array will be in the range of [-10000, 10000].

给定一个长度为2n的数组,要求把数组分成n组,即(a1, b1), (a2, b2), ..., (an, bn) ,使得每组的最小值之和最大。

使用贪心策略,比如样例中,4和3肯定不能和1配对,因为1肯定是最小的,不能拿很大的数和1陪葬,所以只拿2和1配对;然后3、4配对。所以规律就是对数组排序,从小到大每两个配对。那么每组最小值之和就是第0、2、4...个数之和。

代码如下:

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
    	sort(nums.begin(), nums.end());
    	int ans = 0;
    	for(int i = 0; i < nums.size(); i += 2) ans += nums[i];
    	return ans;
    }
};

本代码提交AC,用时92MS。

Leave a Reply

Your email address will not be published. Required fields are marked *