1.自己写：把每一个数拿出来和后面的数挨个对比，时间复杂度n^2

class Solution(object):
    def findDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        j = 0
        k = 0     
        n = len(nums)
        duplication =[]
        while j < n-1:
            for i in range(j+1,n):
                if nums[j] == nums[i]:
                    duplication.append(nums[j])
                    k += 1
                    break
            j += 1
        return duplication