LeetCode.561.Array Partition I 数组拆分 I

题目描述

561 Array Partition I https://leetcode-cn.com/problems/array-partition-i/ https://leetcode.com/problems/array-partition-i/description/

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].

解题过程

2n个整数，两两一对，取每对中较小者之和，使和最大。 最大数肯定取不到，想要取到第二大数就只能把第二大和最大的配成一对，剩下的也依次这么考虑，所以只能是从小到大排序然后取每对中的较小者。 所以方法就是递增排序，然后取奇数位数字之和。

public int arrayPairSum(int[] nums) {
    Arrays.sort(nums);//正序排序
    int sum = 0;
    for (int i = 0; i < nums.length; ) {
        sum += nums[i];
        i = i + 2;
    }
    return sum;
}
}

一次AC，通过这道题还学习了java中Arrays类的使用，java.util.Arrays提供了很多操作数组的方法，以后会经常使用。

GitHub代码

algorithms/leetcode/leetcode/_561_ArrayPartition1.java https://github.com/masikkk/algorithms/blob/master/leetcode/leetcode/_561_ArrayPartition1.java