8.15 - hard - 53

291. Word Pattern II

又是一道backtracking的题目，修修补补也算是AC了

class Solution(object):
    def wordPatternMatch(self, pattern, s):
        """
        :type pattern: str
        :type str: str
        :rtype: bool
        """
        m = {}
        for c in pattern:
            if c not in m:
                m[c] = ""
        return self.dfs(pattern, s, m, 0, 0)
    
    def dfs(self, pattern, s, m, p, pos):
        if pos == len(s) and p == len(pattern):
            return True
        if pos == len(s):
            return False
        if p == len(pattern):
            return False
        
        for i in range(pos, len(s)):
            cur = s[pos:i+1]
            cur_key = None
            for key, value in m.iteritems():
                if value == cur:
                    cur_key = key
            if cur_key and cur_key != pattern[p]:
                continue
            if m[pattern[p]]:
                if m[pattern[p]] == cur:
                    if self.dfs(pattern, s, m, p+1, i+1):
                        return True
            else:
                m[pattern[p]] = cur
                if self.dfs(pattern, s, m, p+1, i+1):
                    return True
                m[pattern[p]] = ""
        return False