Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Example 1:

Input: nums = [-1,0,1,2,-1,-4] Output: [[-1,-1,2],[-1,0,1]] Explanation: nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0. nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0. nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0. The distinct triplets are [-1,0,1] and [-1,-1,2]. Notice that the order of the output and the order of the triplets does not matter. Example 2:

Input: nums = [0,1,1] Output: [] Explanation: The only possible triplet does not sum up to 0. Example 3:

Input: nums = [0,0,0] Output: [[0,0,0]] Explanation: The only possible triplet sums up to 0.

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        answer_list = []
        nums.sort()
        print(nums)
        for index, test_num in enumerate(nums[:-2]):
            if index != 0 and test_num == nums[index-1]:
                continue
            left_index = index + 1
            right_index = len(nums) - 1
            while left_index < right_index:
                left = nums[left_index]
                right = nums[right_index]
                sum = test_num + left + right
                if sum > 0:
                    right_index -= 1
                elif sum < 0:
                    left_index += 1
                elif sum == 0:
                    answer_list.append([test_num, nums[left_index], nums[right_index]])
                    while left == nums[left_index] and left_index < right_index:
                        left_index += 1

        return answer_list


sol = Solution()
print(sol.threeSum([-1,0,1,2,-1,-4]))