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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