LeetCode 561. Array Partition I

原创于 2017-09-08 11:15:05 发布 · 392 阅读

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本文介绍了一种算法问题——数组配对求和。任务是将给定的2n个整数分成n组，使得每组中小的数之和尽可能大。通过排序并选择每对中的较小数来实现最优解。

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Array Partition I

题目描述:

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].

题目大意:

给定2n个元素的数字，让数字两两组合，组成n对，取出每一对中对较小的那个数，求和。这个和最大是多少。
排序后按顺序两两相加求和即可。

题目代码:

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