Go语言题解LeetCode561数组拆分

输入：nums = [1,4,3,2]
输出：4
解释：所有可能的分法（忽略元素顺序）为：
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
所以最大总和为 4

输入：nums = [6,2,6,5,1,2]
输出：9
解释：最优的分法为 (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 n[20001]={0},i=0,j=0,sum=0;
        for(i=0;i<nums.size();i++)
        n[nums[i]+10000]++;        //变相将nums按顺序大小排好，并将顺序存储在n[i]中
        for(i=0;i<20001;)
        {
            if(n[i])
            {
                if(j%2==0) sum+=i-10000;     //按照n[i]中存储的顺序，求得nums[i]
                j++;                  //只有n[i]中有数时，即nums存在这个数时，j才增加，将j变为下角标
                n[i]--;           //此步防止nums中含有两个相同的数
            }
            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])