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3sum.py
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3sum.py
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# Time: O(n^2)
# Space: O(1)
# Given an array S of n integers,
# are there elements a, b, c in S such that a + b + c = 0?
# Find all unique triplets in the array which gives the sum of zero.
#
# Note:
# Elements in a triplet (a,b,c) must be in non-descending order. (ie, a <= b <= c)
# The solution set must not contain duplicate triplets.
# For example, given array S = {-1 0 1 2 -1 -4},
#
# A solution set is:
# (-1, 0, 1)
# (-1, -1, 2)
import collections
class Solution(object):
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
nums, result, i = sorted(nums), [], 0
while i < len(nums) - 2:
if i == 0 or nums[i] != nums[i - 1]:
j, k = i + 1, len(nums) - 1
while j < k:
if nums[i] + nums[j] + nums[k] < 0:
j += 1
elif nums[i] + nums[j] + nums[k] > 0:
k -= 1
else:
result.append([nums[i], nums[j], nums[k]])
j, k = j + 1, k - 1
while j < k and nums[j] == nums[j - 1]:
j += 1
while j < k and nums[k] == nums[k + 1]:
k -= 1
i += 1
return result
def threeSum2(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
d = collections.Counter(nums)
nums_2 = [x[0] for x in d.items() if x[1] > 1]
nums_new = sorted([x[0] for x in d.items()])
rtn = [[0, 0, 0]] if d[0] >= 3 else []
for i, j in enumerate(nums_new):
if j <= 0:
numss2 = nums_new[i + 1:]
for x, y in enumerate(numss2):
if 0 - j - y in [j, y] and 0 - j - y in nums_2:
if sorted([j, y, 0 - j - y]) not in rtn:
rtn.append(sorted([j, y, 0 - j - y]))
if 0 - j - y not in [j, y] and 0 - j - y in nums_new:
if sorted([j, y, 0 - j - y]) not in rtn:
rtn.append(sorted([j, y, 0 - j - y]))
return rtn