561. Array Partition I

QUESTION:

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.

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].

EXPLANATION:

其实我一开始想到的是贪心算法，如果每一步获取到的都是最优的那么总和加起来那么也会是最优的。那么每一步怎么样才能获取到最优的呢，就是ai和bi两者的差是最小的。这样的话算法就显而易见了。

SOLUTION:

public class Solution {
    public int arrayPairSum(int[] nums) {
        Arrays.sort(nums);
        int result = 0;
        for(int i = 0;i<nums.length;i+=2){
            result += nums[i];
        }
        return result;
    }
}
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