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:

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].

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;
}
};