[Leetcode/파이썬] 15. 3Sum

3Sum

---

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.

 

Constraints:

• 3 <= nums.length <= 3000
• -105 <= nums[i] <= 105

 

class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res = set()

        nums.sort()

        for i in range(len(nums) - 2) :
            if i > 0 and nums[i] == nums[i-1] :
                    continue

            j, k = i + 1, len(nums) - 1
            
            while j < k :
                target = nums[i] + nums[j] + nums[k]
                if target == 0 :
                    res.add((nums[i], nums[j], nums[k]))
                    j += 1
                    k -= 1

                elif target > 0 :
                    k -= 1
                else :
                    j += 1

        return [r for r in res]