561. Array Partition I; Difficulty : Easy

最优整数配对求和

最新推荐文章于 2023-11-05 16:31:56 发布

原创最新推荐文章于 2023-11-05 16:31:56 发布 · 464 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LeetCode

LeetCode 专栏收录该内容

23 篇文章

订阅专栏

本文介绍了一种算法问题，即如何从2n个整数中找出n组配对，使得每组配对中较小数值之和尽可能大。通过排序和选择偶数位置元素的方法给出了解决方案，并提供了C语言及Python实现。

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].

算法分析：只要把相差最小的两个数分成一组即可。

C语言版

int cmp(const void *a, const void *b)
{
    return *(int *)a - *(int *)b;
}

int arrayPairSum(int* nums, int numsSize) {
    int i, sum = 0;
    qsort(nums, numsSize, sizeof(int), cmp);
    for(i = 0; i < numsSize; i += 2)
        sum += nums[i];

    return sum;
}

Python版

class Solution(object):
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return sum(sorted(nums)[0:len(nums):2])