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: 4
Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 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].

思考

将给出的2n个整数两个一组，使得每组的较小值的和最大。通过观察，可以发现将相近的两个整数放在一组的结果是最大的，因为这样组合所“损失”的差是最少的，比如Example 1中的例子：
输入为：[1, 4, 3, 2]
分组方式一共有3种

分组方式 (1, 2)(3, 4) (1, 3)(2, 4) (1, 4)(2, 3) 输出结果 4 3 3

因此，可以先排序，然后将第2n + 1个(n = 0， 1， 2…)的数值（即每组的较小值）加起来则为最大结果。

我的答案

c++

class Solution {public:    int arrayPairSum(vector<int>& nums) {        sort(nums.begin(), nums.end());        int result = 0;        for (int i = 0; i < nums.size(); i = i + 2) {            result += nums[i];        }        return result;    }};