我怎样才能使这个 3sum 解决方案更省时？

Question

I came up with a solution for the 3sum problem on leetcode but even though it passes all test codes it fails because of 'time limit exceeded'.

class Solution(object):
def threeSum(self, nums):
    triplet_array = []
    
    x = []
    for i in range(len(nums)):
        for j in range(i+1, len(nums)):
            for k in range(j+1, len(nums)):
                if nums[i] + nums[j] + nums[k] == 0:
                    if i !=j and i !=k and j != k:
                                if [nums[i], nums[j], nums[k]] != x:
                                    x = [nums[i], nums[j], nums[k]]
                                    x.sort()
                                    if x not in triplet_array:
                                        triplet_array.append(x)
    return triplet_array

I don't know how to perform the 'space/time' analysis that it keeps referencing so I'm wondering how I go about making this solution more efficient so that it will pass all test cases in the allotted time.

For reference the 3sum problem details can be found here: https://leetcode.com/problems/3sum/

Answer 1

One tip would be to sort the incoming array first. If you know that three elements at indices i, j, and k add up to more than 0, then increasing i, j, or k (so long as the list is sorted) will also be more than 0.