561. Array Partition I

Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a₁, b₁), (a₂, b₂), ..., (a_n, b_n) which makes sum of min(a_i, b_i) 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].

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给一个2n大小的序列，也就是有n对数，取出每对数中的最小值求和，问最大值是多少。想法是对于一个小的数，要拿一个刚好比他大的数和他组队，不然就浪费一个较大的数。所以实现是先排序，然后每隔一个数取一个数相加得到结果。

代码：

class Solution 
{
public:
	int arrayPairSum(vector<int>& nums) 
	{
		int res = 0;
		sort(nums.begin(), nums.end());
		for(int i = 0; i < nums.size(); i+=2)
		{
			res += nums[i];
		}
		return res;
	}
};