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 Patternclass 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 Trueif not pattern or not str:return Falseif len(pattern)>len(str):return Falsep = pattern[0]if p in pattern_map:if str[:len(pattern_map[p])]!=pattern_map[p]:return Falseelse: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 Truedel pattern_map[p]return Falsereturn 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"