561. Array Partition I(Java)

最新推荐文章于 2022-01-31 22:01:47 发布

Crystal_ting

最新推荐文章于 2022-01-31 22:01:47 发布

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分类专栏： leetcode Array

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本文介绍了一道算法题目，任务是将给定的2n个整数分成n组，每组两个整数，使得所有组中小的那个整数之和尽可能大。通过排序和选择每个偶数位置上的元素来解决这个问题。

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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].

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