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题目

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).

题目看了半天才看懂。。。将数组分为两个两个一组，使得所有小组中的最小数之和最大

我的想法

按照大小排序，让两个相邻数为一组，此时两个数之间的差值最小

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

解答

一样的思路

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