Array Partition I

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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].

思路：临近的值相互比较，这样可以min掉最小的差值，最后的合才是最大。

class Solution {
    public int arrayPairSum(int[] nums) {
        if(nums == null || nums.length ==0) return 0;
        Arrays.sort(nums);
        int res = 0;
        for(int i=0; i<nums.length/2; i++){
            res += Math.min(nums[2*i], nums[2*i+1]);
        }
        return res;
    }
}