Leetcode 561: Array Partition I

题目传送门:
Leetcode 561. Array Partition I

题目:

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: 4Explanation: 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].

思路:

我觉得 solution 第一个解释的非常好…

1. Assume in each pair i, bi >= ai.
2. 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.
3. Denote Sa = a1 + b1 + a2 + b2 + … + an + bn. Sa is constant for a given input.
4. Denote di = |ai - bi|. Given 1, di = bi - ai. Denote Sd = d1 + d2 + … + dn.
5. So Sa = a1 + a1 + d1 + a2 + a2 + d2 + … + an + an + di = 2Sm + Sd => Sm = (Sa - Sd) / 2. To get the max Sm, given Sa is constant, we need to make Sd as small as possible.
6. 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.

中文总结:
对连续的数进行分组,可得到最大的sum和.

不值一提的代码

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