leetcode 561:Array Partition I

原题：

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: 4Explanation: 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].

就是两两配对，去最小值，加起来最大。

代码如下：

int arrayPairSum(int* nums, int numsSize) {    int comp(const void* a,const void* b)    {        return *(int*)a-*(int*)b;    }    qsort(nums,numsSize,sizeof(int),comp);    int result=0;    for(int n=0;n<numsSize;n+=2)    {        result+=*(nums+n);    }    return result;}

排序以后跳着加。

时间就是O(n)。