[LeetCode] 0015. 3Sum

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內容目錄

Medium

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

Python

				
					class Solution:
    def twoSum(self, nums: List[int], i: int, result: List[List[int]]):
        low = i + 1
        hight = len(nums) - 1
        while (low < hight):
            sum = nums[i] + nums[low] + nums[hight]
            if sum < 0:
                low += 1
            elif sum > 0:
                hight -= 1
            else:
                result.append([nums[i], nums[low], nums[hight]])
                low += 1
                hight -= 1
                while low < hight and nums[low] == nums[low - 1]:
                    low += 1

    def threeSum(self, nums: List[int]) -> List[List[int]]:
        result = []
        nums.sort()
        for i, item in enumerate(nums):
            if item > 0:
                break
            if item != nums[i-1] or i == 0:
                self.twoSum(nums, i, result)

        return result

Medium

Python

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