palindrome_partitioning

View Source

 1# @leet start
 2class Solution:
 3    def partition(self, s: str) -> list[list[str]]:
 4        def is_palindrome(s):
 5            return s == s[::-1]
 6
 7        def dfs(s, path, result):
 8            if not s:
 9                result.append(path)
10                return
11            for i in range(1, len(s) + 1):
12                if is_palindrome(s[:i]):
13                    dfs(s[i:], path + [s[:i]], result)
14
15        result = []
16        dfs(s, [], result)
17        return result
18
19
20# @leet end
21
22
23def test():
24    assert 2 + 2 == 4

class Solution: View Source

 3class Solution:
 4    def partition(self, s: str) -> list[list[str]]:
 5        def is_palindrome(s):
 6            return s == s[::-1]
 7
 8        def dfs(s, path, result):
 9            if not s:
10                result.append(path)
11                return
12            for i in range(1, len(s) + 1):
13                if is_palindrome(s[:i]):
14                    dfs(s[i:], path + [s[:i]], result)
15
16        result = []
17        dfs(s, [], result)
18        return result

def partition(self, s: str) -> list[list[str]]: View Source

 4    def partition(self, s: str) -> list[list[str]]:
 5        def is_palindrome(s):
 6            return s == s[::-1]
 7
 8        def dfs(s, path, result):
 9            if not s:
10                result.append(path)
11                return
12            for i in range(1, len(s) + 1):
13                if is_palindrome(s[:i]):
14                    dfs(s[i:], path + [s[:i]], result)
15
16        result = []
17        dfs(s, [], result)
18        return result

def test(): View Source

24def test():
25    assert 2 + 2 == 4