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Palindrome Partitioning @ LeetCode (Python)

kitt posted @ 2014年4月09日 08:53 in LeetCode , 3226 阅读

先求出每个子串是否为palindrome, 用isPal记录。isPal[i][j] == true含义是从s[i]到s[j]的子串(包含起点终点)是回文串。再用DFS。

First use isPal to record each substring is palindrome or not. "isPal[i][j] == true" means a substring starts from s[i] (included) to s[j] (included) is a palindrome. Then use DFS.

class Solution:
    # @param s, a string
    # @return a list of lists of string
    def partition(self, s):
        self.s, self.res, self.lenS =s, [], len(s)
        self.isPal = [ [False for j in xrange(self.lenS)] for i in xrange(self.lenS) ]
        for j in xrange(self.lenS):
            for i in reversed(xrange(j + 1)):
                if s[i] == s[j] and ( j - i <= 1 or self.isPal[i + 1][j - 1] == True): 
                    self.isPal[i][j] = True # True means substring from s[i](included) to s[j](included) is palindrome
        self.dfs(0, [])
        return self.res
    
    def dfs(self, start, L):
        if start == self.lenS: 
            self.res.append(L)
            return
        for end in xrange(start, self.lenS):
            if self.isPal[start][end] == True:
                self.dfs( end + 1, L[:] + [self.s[start : end + 1]] ) # make a copy of list L and add a substring

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