题目描述

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 = min(1, 2) + min(3, 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].

解法一

思路

这题看起来有点绕,理清楚了就会发现很简单了。

首先要明确一点,最后是需要在2n个数里面找出n个(不重复)的数,即每个数用一次。再来看“结果最大”,所以我们希望找到的n个数尽可能大。那么假设2n个数是有序的,考虑最后一个数,这个数最大,但却不可能在结果中,因为min(a, b)永远都取不到最大的数。那再来看看倒数第二大的数,为了让这个数能够在结果中,我们就只能找一个比它更大的数来匹配,而显然这个数只能是最后那个最大的数。顺着这个思路,就会发现要取到的数一定是数组排序后,偶数index的数,最后全部相加就是最终结果啦。

Python

class Solution(object):
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nums.sort()
        return sum(nums[::2])

Java

public class Solution {
    public int arrayPairSum(int[] nums) {
        Arrays.sort(nums);
        int sum = 0;
        for (int i = 0; i < nums.length; i += 2) {
            sum += nums[i];
        }
        return sum;
    }
}

C++

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