【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.

解析：

数组分割1
给一组2n个整数，分割为n组，一组2个数，使min(a,b)之和最大
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).

我的解法

很明显每组数值越接近，浪费的数值越小，所得的总和最大，我的解法如下

public class ArrayPartitionI {
    public static int arrayPartitionI(int[] array) {
        Arrays.sort(array);
        int sum = 0;
        for (int i = 0; i < array.length - 1; i += 2) {
            sum += array[i];
        }
        return sum;
    }
}

最优解法

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;
    }
}

哈哈几乎一样，后来想了一下不用length-1也可以的