def threeSum(nums):

         """

         :type nums: List[int]

         :rtype: List[List[int]]

         """

         def twoSum(a,b,c):

             if a + b + c == 0:

                 return True

             else:

                 return False

         res = []

         for i in range(len(nums)-2):

             a = nums[i]

             for j in range(i+1,len(nums)-1):

                 b = nums[j]

                 k = j +1

                 #print(i,j,k,a,b,nums[k])

                 while k < len(nums):

                     print(a,b,nums[k])

                     if twoSum(a,b,nums[k]) :

                         print('....',a,b,nums[k])

                         temp = [a,b,nums[k]]

                         temp.sort()

                         print('...temp',temp)

                         if temp not in res:

                             res.append(temp)

                         print('....res',res)

                     k += 1

         return res

 class Solution(object):

     '''

     O(n^2)时间复杂度

     '''

     def threeSum(self, nums):

         """

         :type nums: List[int]

         :rtype: List[List[int]]

         """

         result = []

         num_cnt_dict = {}

         for num in nums:

             num_cnt_dict[num] = num_cnt_dict.get(num, 0) + 1

         if 0 in num_cnt_dict and num_cnt_dict[0] >= 3:

             result.append([0, 0, 0])

         nums = sorted(list(num_cnt_dict.keys()))

         for i, num in enumerate(nums):

             for j in nums[i+1:]:

                 if num * 2 + j == 0 and num_cnt_dict[num] >=2:

                     result.append([num, num, j])

                 if j * 2 + num == 0 and num_cnt_dict[j] >= 2:

                     result.append([j, j, num])

                 diff = 0 - num - j

                 if diff > j and diff in num_cnt_dict:

                     result.append([num, j, diff])

         return result

                if diff > b and diff in nums:  #要注意>b，才能不和前面的if重合

                    res.append([a,b,diff])

        return res

 class Solution(object):

     '''

     O(n^2)时间复杂度

     '''

     def threeSum(self, nums):

         """

         :type nums: List[int]

         :rtype: List[List[int]]

         """

         res = []

         dic = {}

         for item in nums:

             dic[item] = dic.get(item,0) + 1

         if 0 in dic and dic[0] >= 3:

             res.append([0,0,0]) #为什么这个要单独列出来 -》因为除了这种情况，for循环中不会再出现三个数相等且和为0的情况

         #nums = sorted(list(dic.keys()))

         neg = list(filter(lambda x:x<0,dic.keys()))

         pos = list(filter(lambda x:x>=0,dic.keys()))

         for a in neg:

             for b in pos:#为什么j的有边界不是len(nums)-1 -》因为可能会存在第三个数巧合和nums[j]相等

                 diff  = 0 - a - b

                 if diff in dic:

                     if diff in (a,b) and dic[diff] >= 2:

                         res.append([a,b,diff])

                     if diff<a or diff>b:

                         res.append([a,b,diff])

         return res