[LeetCode]561. Array Partition I (数组分区 1)

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), …, (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.

题目大意：
给定一个长度为2n(偶数)的数组，分成n个小组，返回每组中较小值的和sum，使sum尽量大
Example 1:

Input: [1,4,3,2]
Output: 4
Explanation: n is 2, and the maximum sum of pairs is 4.

Note:

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].

思路：

先排序，将相邻两个数分为一组，每组较小数都在左边，求和即可

算法分析：
查看英文版请点击上方

假设对于每一对i，bi >= ai。
定义Sm = min（a1，b1）+ min（a2，b2）+ … + min（an，bn）。最大的Sm是这个问题的答案。由于bi >= ai，Sm = a1 + a2 + … + an。
定义Sa = a1 + b1 + a2 + b2 + … + an + bn。对于给定的输入，Sa是常数。
定义di = | ai - bi |。由于bi >= ai，di = bi-ai, bi = ai+di。
定义Sd = d1 + d2 + … + dn。
所以Sa = a1 + (a1 + d1) + a2 + (a2 + d2) + … + an + (an + di) = 2Sm + Sd , 所以Sm =（Sa-Sd）/ 2。为得到最大Sm，给定Sa为常数，需要使Sd尽可能小。
所以这个问题就是在数组中找到使di（ai和bi之间的距离）的和尽可能小的对。显然，相邻元素的这些距离之和是最小的。

代码如下：

// https://leetcode.com/problems/array-partition-i/#/description原题
//https://discuss.leetcode.com/topic/87206/java-solution-sorting-and-rough-proof-of-algorithm
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:int arrayPairSum(vector<int>& nums) {int res = 0;sort(nums.begin(), nums.end());for(int i=0; i<nums.size(); i+=2){res += nums[i];}return res;}
};
int main()
{Solution a;int num[4] = {1, 4, 3, 2};int numLength = sizeof(num) / sizeof(num[0]);vector<int> nums(num, num+numLength);cout << a.arrayPairSum(nums) << endl;return 0;
}

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