561. Array Partition I

置顶

Created At : 2023-10-28 15:14

Count:313 Views 👀 :

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:

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].

这道题让我们分割数组，两两一对，让每对中较小的数的和最大。这题难度不大，用贪婪算法就可以了。由于我们要最大化每对中的较小值之和，那么肯定是每对中两个数字大小越接近越好，因为如果差距过大，而我们只取较小的数字，那么大数字就浪费掉了。明白了这一点，我们只需要给数组排个序，然后按顺序的每两个就是一对，我们取出每对中的第一个数即为较小值累加起来即可，参见代码如下：

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;
    }
};

LeetCode All in One 题目讲解汇总(持续更新中…)

转载请注明来源于 Grandyang 的博客 (grandyang.com)，欢迎对文章中的引用来源进行考证，欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论，也可以邮件至 grandyang@qq.com