边工作边刷题：70天一遍leetcode: day 86

Word Pattern II

要点：

• 注意与I的差异，其实题不难，看到这种迷乱的，首先要想到backtrack
• 1:1 mapping两个条件：p in and str in, or p not in and str not in values().

错误点：

• false的退出条件是pattern没有足够字符1:1了（i.e., len(pattern)>len(str)），但true的条件是pattern和str都比完了

https://repl.it/CekC/1

# Given a pattern and a string str, find if str follows the same pattern.

# Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty substring in str.

# Examples:
# pattern = "abab", str = "redblueredblue" should return true.
# pattern = "aaaa", str = "asdasdasdasd" should return true.
# pattern = "aabb", str = "xyzabcxzyabc" should return false.
# Notes:
# You may assume both pattern and str contains only lowercase letters.

# Hide Company Tags Dropbox Uber
# Hide Tags Backtracking
# Hide Similar Problems (E) Word Pattern

class Solution(object):
    def wordPatternMatch(self, pattern, str):
        """
        :type pattern: str
        :type str: str
        :rtype: bool
        """
        pattern_map = {}
        def helper(pattern, str):
            if not pattern and not str:
                return True

            if not pattern or not str:
                return False

            if len(pattern)>len(str):
                return False

            p = pattern[0]
            if p in pattern_map:
                if str[:len(pattern_map[p])]!=pattern_map[p]:
                    return False
                else:
                    return helper(pattern[1:], str[len(pattern_map[p]):])

            for i in xrange(len(str)):
                substr = str[:i+1]
                if substr not in pattern_map.values():
                    pattern_map[p]=str[:i+1]
                    if helper(pattern[1:], str[i+1:]):
                        return True
                    del pattern_map[p]

            return False

        return helper(pattern, str)

sol = Solution()
assert sol.wordPatternMatch("abab", "redblueredblue")==True, "must be True"
assert sol.wordPatternMatch("aaaa", "asdasdasdasd")==True, "must be True"
assert sol.wordPatternMatch("aabb", "xyzabcxzyabc")==False, "must be False"