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threeSum.py
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94 lines (80 loc) · 2.79 KB
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from collections import Counter
from typing import List
class Solution(object):
# 혼자 풀이(런타임 에러)
def threeSum(self, nums):
answer = []
# 3개의 값의 합이므로 len(nums) < 3이면 빈 값 return
if len(nums) < 3:
return answer
if max(nums) == 0:
answer.append([0,0,0])
return answer
# nums 배열 오름차순 정렬
nums = sorted(nums)
# len(nums) 만큼 for문 수행
for i in range(len(nums)):
for j in range(i+1,len(nums)):
for k in range(j + 1, len(nums)):
if nums[i]+nums[j]+nums[k] == 0:
array = []
array.append(nums[i])
array.append(nums[j])
array.append(nums[k])
answer.append(array)
if len(answer) == 1:
return answer
answer = sorted(answer)
an = []
print(answer)
for ans in range(len(answer)):
dic = Counter(answer[ans])
if dic != Counter(answer[ans+1]):
an.append(answer[ans])
else:
if ans+1 == len(answer)-1:
an.append(answer[ans])
break;
return an
# 책 풀이
def threeSum_book(self, nums: List[int]) -> List[List[int]]:
results = []
nums.sort()
for i in range(len(nums) - 2):
# 중복된 값 건너뛰기
if i > 0 and nums[i] == nums[i - 1]:
continue
# 간격을 좁혀가며 합 `sum` 계산
left, right = i + 1, len(nums) - 1
while left < right:
sum = nums[i] + nums[left] + nums[right]
if sum < 0:
left += 1
elif sum > 0:
right -= 1
else:
# `sum = 0`인 경우이므로 정답 및 스킵 처리
results.append([nums[i], nums[left], nums[right]])
while left < right and nums[left] == nums[left + 1]:
left += 1
while left < right and nums[right] == nums[right - 1]:
right -= 1
left += 1
right -= 1
return results
if __name__ == "__main__":
solution = Solution()
nums = [-1,0,1,2,-1,-4]
print(solution.threeSum(nums))
nums = [0, 0, 0]
print(solution.threeSum(nums))
nums = [1,1,-2]
print(solution.threeSum(nums))
nums = [-2,0,0,2,2]
print(solution.threeSum(nums))
nums = [-1, 0, 1]
print(solution.threeSum(nums))
nums = [0, 0, -1]
print(solution.threeSum(nums))
nums = [-2,0,1,1,2]
print(solution.threeSum(nums))