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].
这题很简单的啊。

由于自己Java没学过，还需要慢慢的积累类似Arrays.sort()这些函数的用法。

class Solution{
public int arrayPairSum(int [] nums) {
Arrays.sort(nums);  //这里直接用相关的模块求排序就行
int sum=0;
for (int i=0;i<(nums.length/2);i++) {
sum+=nums[2*i];
}
return sum;
}

}

class Solution:

    def array
4000
PairSum(self, nums):

        """

        :type nums: List[int]

        :rtype: int

        """

        sort_nums=sorted(nums)

        sum=0

        n=int(len(nums)/2)

        print(n)

        for i in range(n):

            sum+=sort_nums[2*i]

        return sum