[每日LeetCode] 561. Array Partition I

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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 = min(1, 2) + min(3, 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].

思路:本题要求对数组中的元素两两分组,取每组的最小值,最后把每组最小值相加,求如何组合得到最大的和。细想本题,我们只需使每个组的两个元素相差尽量小,以免浪费较大的数字。先对数组排序,从第一个元素开始,隔一个元素取一个数字,再相加即可。


C++代码

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

运行时间:92ms

运行内存:11.4M

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