【跟Leon一起刷LeetCode】561. Array Partition I

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

[b]Example 1：[/b]

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).

[b]Note：[/b]

n is a positive integer, which is in the range of [1, 10000].

All the integers in the array will be in the range of [-10000, 10000].

[b]问题描述:[/b]

将长度为2n的数组分为n组整数，每组2个元素称为一个pair，将所有pair中较小的那个元素取出来,并求和。如何使得求和结果最大?

Solution：

题目比较简单，如何划分这个pair。主要是对数组进行排序，之后两两划分。

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

在

Discuss

上面看到高人提出了更好的算法：

Assume in each pair i, bi >= ai.

Denote Sm = min(a1, b1) + min(a2, b2) + … + min(an, bn). The biggest Sm is the answer of this problem. Given 1, Sm = a1 + a2 + … + an.

Denote Sa = a1 + b1 + a2 + b2 + … + an + bn. Sa is constant for a given input.

Denote di = |ai - bi|. Given 1, di = bi - ai. Denote Sd = d1 + d2 + … + dn.

So Sa = a1 + a1 + d1 + a2 + a2 + d2 + … + an + an + dn = 2Sm + Sd => Sm = (Sa - Sd) / 2. To get the max Sm, given Sa is constant, we need to make Sd as small as possible.

So this problem becomes finding pairs in an array that makes sum of di (distance between ai and bi) as small as possible. Apparently, sum of these distances of adjacent elements is the smallest. If that’s not intuitive enough, see attached picture. Case 1 has the smallest Sd.