Palindrome Partitioning II 解答

Question

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

Example:

Input: "aab" Output: 1 Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.

Solution

第一个思路是和Palindrome Partitioning解答思路相似，先用二维DP得到palindrome结果，然后用DFS.

时间复杂度O(N^2) 空间复杂度O(N^2)

 1 class Solution:
 2     def check(self, s: str) -> None:
 3         n = len(s)
 4         self.dp = [[False for i in range(n)] for j in range(n)]
 5         for i in range(n):
 6             self.dp[i][i] = True
 7         for i in range(n - 1):
 8             if s[i] == s[i + 1]:
 9                 self.dp[i][i + 1] = True
10         for k in range(3, n + 1):
11             for i in range(n - k + 1):
12                 j = i + k - 1
13                 if s[i] == s[j] and self.dp[i + 1][j - 1]:
14                     self.dp[i][j] = True
15     
16     def dfs(self, s: str, start: int, record: List[str], result: List[int]) -> None:
17         n = len(s)
18         if start == n:
19             cut_num = len(record) - 1
20             result[0] = min(cut_num, result[0])
21             return
22         for end in range(start, n):
23             if self.dp[start][end]:
24                 record.append(s[start: end + 1])
25                 self.dfs(s, end + 1, record, result)
26                 record.pop()
27     
28     def minCut(self, s: str) -> int:
29         self.check(s)
30         result = [len(s) - 1]
31         self.dfs(s, 0, [], result)
32         return result[0]
33