[Leetcode] 561. Array Partition I 解题报告

原创于 2017-12-24 09:35:37 发布 · 267 阅读

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本文介绍了一种算法问题的解决方案，即如何通过将给定的2n个整数分成n组，使得每组中较小数值之和尽可能大。通过对整数数组进行排序并选取每个偶数位置上的元素来实现这一目标。

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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].

思路：

练手题目，哈哈。首先对nums排序，然后将其奇数索引下的数加起来即可。

代码：

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