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LeetCode: 131. Palindrome Partitioning 본문
class Solution:
def partition(self, s: str) -> List[List[str]]:
self.ret = []
self.dfs(s, [])
return self.ret
def dfs(self, s, path):
if not s:
self.ret.append(path)
return
for i in range(1, len(s)+1):
first, second = s[:i], s[i:]
if self.is_Palindromic(first):
self.dfs(second, path+[first])
def is_Palindromic(self, s):
return s == s[::-1]'Computer Science > Coding Test' 카테고리의 다른 글
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| LeetCode: 132. Palindrome Partitioning II (0) | 2021.11.13 |
| LeetCode: 146. LRU Cache (0) | 2021.11.11 |
| LeetCode: 130. Surrounded Regions (0) | 2021.11.10 |
| LeetCode: 129. Sum Root to Leaf Numbers (0) | 2021.11.10 |