#Leetcode# 561. Array Partition I

https://leetcode.com/problems/array-partition-i/

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].

代码:

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
        int n = nums.size();
        sort(nums.begin(), nums.end());
        int sum = 0;
        for(int i = 0; i < n; i += 2) {
            sum += min(nums[i], nums[i + 1]);
        }
        return sum;
    }
};

  按照 submit 之后推荐的题目做就变成了从链表到数组

FHFHFH 

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转载自www.cnblogs.com/zlrrrr/p/10291978.html