Leetcode Third Maximum Number problem solution

class Solution:
    def thirdMax(self, nums: List[int]) -> int:
        d=len(set(nums))
        x=sorted(set(nums))
        if d==2:
            return x[-1]
        elif d==1:
            return nums[0]
        else:
            return x[-3]

public int thirdMax(int[] nums) 
    {
        int keepMax = nums[0];
        int thirdMax = keepMax;
        
        long max = Long.MAX_VALUE;
        for (int i = 0; i < 3; i++)
        {
            long tmp = Long.MIN_VALUE;
            for (int j = 0; j < nums.length; j++)
            {
                if (nums[j] < max && nums[j] > tmp) tmp = nums[j];
            }
            if (i == 0) { keepMax = (int)tmp; max = tmp; }

            if (i == 1) { max = tmp;}

            if (i == 2)
            {   
                if (tmp == Long.MIN_VALUE) return keepMax;
                else thirdMax = (int)tmp;
            }
        }
        
        return thirdMax;
    }

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        set<int>S(nums.begin(),nums.end());
        priority_queue<int> pq(S.begin(),S.end());
        if(pq.size() == 1 || pq.size() == 2) return pq.top();
        pq.pop();
        pq.pop();
        return pq.top();
    }
 };

int j=0;
	int length=0;
	int m=0;
   int* r= malloc (sizeof(int)*numsSize) ;
for(int i=0;i<numsSize-1;i++)
{
	for( j=0;j<numsSize-1-i;j++)
	{
		if(nums[j]<nums[j+1])
		{
			int temp = nums[j];
			nums[j]=nums[j+1];
			nums[j+1]=temp;
		}
	}
}



for (int i=0; i<numsSize; i++)
{
for (j=0; j<i; j++)
{
if (nums[i] == nums[j])
break;
}
if (i == j)
{
	r[m++]=nums[i];
	length++;
		}

}

if(length>=3)
    return r[2];
else
    return r[0];