15. 3Sum
Updated: 2024-03-12
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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.
First accepted
Idea:
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
x = 0 # index
ll = len(nums)
res = []
while x < ll - 2:
if x == 0 or nums[x] != nums[x-1]:
y = x + 1
z = ll - 1
while y < z:
s = nums[x] + nums[y] + nums[z]
if s == 0:
res.append([nums[x], nums[y], nums[z]])
while y < z and nums[y] == nums[y+1]:
y += 1
while z > y and nums[z] == nums[z-1]:
z -= 1
y += 1
z -= 1
elif s > 0:
z -= 1
else:
y += 1
x += 1
return res
