[LeetCode] 131. Palindrome Partitioning_Medium tag: DFS, backtracking, Palindrome-蒲公英云

[LeetCode] 131. Palindrome Partitioning_Medium tag: DFS, backtracking, Palindrome

清疚 2021-11-16 07:20 314阅读 0赞

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

Return all possible palindrome partitioning of s.

Example:

Input: "aab"
Output:
[
  ["aa","b"],
  ["a","a","b"]
]

这个题目我们可以理解为是s中不断决定是否要在该character的后面切一刀，可以切，可以不切， 2^n 种可能，让我们想到了subsets[LeetCode] 78. Subsets tag: backtracking，实际上就可以利用subsets的做法，只是temp存储的是s的index，然后加上每切一刀的时候都要判断是否是palindrome，我们利用[Palindrome] Check any substring in a s is a palindrome or not. 来去优化时间复杂度。

T: O(2^n) 因为我们优化了palindrome substring。

Code

class Solution:
    def palindromePartition(self, s):
        ans, palin = [], self.generatePalin(s)
        self.helper(s, [], 0, palin, ans)
        return ans
    def generatePalin(self, s):
        n = len(s)
        palin = [[False] * n for  _ in range(n)]
        for i in range(n):
            palin[i][i] = True
            if i and s[i] == s[i - 1]:
                palin[i - 1][i] = True
        for length in range(2, n):
            for start in range(n):
                if start + length < n and s[start] == s[start + length]:
                    palin[start][start + length] = palin[start + 1][start + length - 1]
        return palin
    def helper(self, s, temp, pos, palin, ans):
        if pos == len(s):
            ans.append(temp)
        for i in range(pos, len(s)):
            if not palin[pos][i]:
                continue
            self.helper(s, temp + [s[pos: i + 1]], i + 1, palin, ans)