輸入：nums = [1,4,3,2]
輸出：4
解釋?zhuān)核锌赡艿姆址ǎê雎栽仨樞颍椋?
1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
所以最大總和為 4

輸入：nums = [6,2,6,5,1,2]
輸出：9
解釋?zhuān)鹤顑?yōu)的分法為 (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9

class Solution {
public:
    int arrayPairSum(vector<int>& nums) {
       /*
        int arr[nums.size()]={0};
        int temp=0;
        for(int i=0;i<nums.size();i++)
        arr[i]=nums[i];
        for(int i=0;i<nums.size()-1;i++)
        {
            for(int j=i+1;j<nums.size();j++)
            if(arr[i]>arr[j]) {temp=arr[i];arr[i]=arr[j];arr[j]=temp;}   //冒泡排序會(huì)超出時(shí)間限制
        }
        int res=0;
        for(int i=0;i<nums.size();i+=2){res+=arr[i];}
        return res;*/
        int n[20001]={0},i=0,j=0,sum=0;
        for(i=0;i<nums.size();i++)
        n[nums[i]+10000]++;        //變相將nums按順序大小排好，并將順序存儲(chǔ)在n[i]中
        for(i=0;i<20001;)
        {
            if(n[i])
            {
                if(j%2==0) sum+=i-10000;     //按照n[i]中存儲(chǔ)的順序，求得nums[i]
                j++;                  //只有n[i]中有數(shù)時(shí)，即nums存在這個(gè)數(shù)時(shí)，j才增加，將j變?yōu)橄陆菢?biāo)
                n[i]--;           //此步防止nums中含有兩個(gè)相同的數(shù)
            }
            else i++;
        }
        return sum;
    }
};

class Solution(object):
    def arrayPairSum(self, nums):
        nums.sort()
        res = 0
        for i in range(0, len(nums), 2):
            res += min(nums[i], nums[i + 1])
        return res

class Solution(object):
    def arrayPairSum(self, nums):
        nums.sort()
        return sum(nums[::2])

class Solution(object):
    def arrayPairSum(self, nums):
        return sum(sorted(nums)[::2])