561. Array Partition I; Difficulty : Easy

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])