561. Array Partition I

xiaoxiao2021-02-28  69

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

转载请注明原文地址: https://www.6miu.com/read-60070.html

最新回复(0)