LeetCode 561 : ArrayPartition I

LeetCode 561 : ArrayPartition I

题目大意

将一个长度为2n的数组分成n对，获取每对中的最小值，使最小值的和最大。

• 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. 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].

• Example

Input: [1,4,3,2]Output: 4Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).

思路分析

将数组从小到大排序，将偶数位的值相加即可。贪心算法，每次都获得当前可获得的最大的最小值，即仅比最大值小的值，然后去除当前最大值，继续比较。

相应代码

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