Array Partition I–LeetCode#561

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.

(给一个数量为2n的整数数组，分成n组，求每组最小值的和(尽可能大))

Example:

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).

思路：将数组递增排序，最大的和是第1，3，5，7…n-1数的和.

public class ArrayPartitionI {
    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;
    }
    public static void main(String[] args) {
        ArrayPartitionI arrayPartitionI=new ArrayPartitionI();
        int[] nums={1,1,1,1};
        System.out.println(arrayPartitionI.arrayPairSum(nums));
    }
}