leetcode 561. 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].

题目大意:给定一个2n长度的整数数组,将其分n对整数, 即(a1, b1), (a2, b2), ..., (an, bn), 求

                                                                                                                                       

 而

                                                                    

 我们有

由于前面那项是常数,我们只要考虑让后面那项最小,即

显然,要让和最小,尽量让每一对数,尽可能接近。经过整理,得到最大值为数组排序后的奇数项之和。

 1 class Solution {
 2 public:
 3     int arrayPairSum(vector<int>& nums) {
 4         sort(nums.begin(), nums.end());
 5         int sum = 0;
 6         for (int i = 0; i < nums.size(); i += 2) {
 7             sum += nums[i];
 8         }
 9         return sum;
10     }
11 };
原文地址:https://www.cnblogs.com/qinduanyinghua/p/11850775.html