LeetCode 561. 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.

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].

Solution

题意要求分割数组，两两一对，使得每对中较小的数的和最大。使用贪婪算法，要最大化每对中的较小值之和，那么每对中两个数字大小越接近越好，因为如果差距过大，而只取较小的数字，那么大数字就浪费了。所以只需给数组排序，然后按顺序的每两个就是一对，取出每对中的第一个数即为较小值，累加起来即可。

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