LeetCode[561]Array Partition I

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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].

本地执行代码：

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

/*
解题思路：遍历向量，每次取出相邻的两个元素，然后比大小，对小的求和
*/

int arrayPairSum(vector<int>& nums);
int getmin(int x, int y);

int main()
{
	vector<int> a(10, 1);
	int b = 0;
	b = arrayPairSum(a);
	cout << b << endl;
	system("pause");
	return 0;
}

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

提交代码：

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