LeetCode 题解(271) : Word Pattern II

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题目：

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 inpattern 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.

题解：

比较典型的Backtracing，注意更新hashmap的时候不要犯错。

C++版：

class Solution {public:    bool wordPatternMatch(string pattern, string str) {        unordered_map<char, string> d;        unordered_map<string, bool> used;        return match(pattern, str, d, used);    }        bool match(string pattern, string str, unordered_map<char, string> &d, unordered_map<string, bool> &used) {        if(pattern.length() == 0 && str.length() == 0)            return true;        if(pattern.length() == 0 || str.length() == 0)            return false;                    if(pattern.length() == 1) {            if(d.find(pattern[0]) == d.end()) {                if(used.find(str) == used.end())                    return true;                else                    return false;            } else {                if(d[pattern[0]] == str)                    return true;                else                    return false;            }        }                if(d.find(pattern[0]) != d.end()) {            if(str.find(d[pattern[0]]) != 0)                return false;            else {                bool result = match(pattern.substr(1), str.substr(d[pattern[0]].length()), d, used);                if(result)                    return true;            }        } else {            int length = str.length() - (pattern.length() - 1);            for(int i = 1; i <= length; i++) {                string cur = str.substr(0, i);                d.insert(pair<char, string>(pattern[0], cur));                used.insert(pair<string, bool>(cur, true));                bool result = match(pattern.substr(1), str.substr(cur.length()), d, used);                d.erase(pattern[0]);                used.erase(cur);                if(result)                    return true;            }        }        return false;    }};

Python版：

class Solution(object):    def wordPatternMatch(self, pattern, str):        """        :type pattern: str        :type str: str        :rtype: bool        """        d, used = {}, {}        return self.match(pattern, str, d, used)                    def match(self, pattern, str, d, used):        if len(pattern) == 0 and len(str) == 0:            return True        elif len(pattern) == 0 or len(str) == 0:            return False                    if len(pattern) == 1:            if pattern in d:                if d[pattern] == str:                    return True                else:                    return False            else:                if str in used and used[str]:                    return False                else:                    return True                if pattern[0] in d:            if str.find(d[pattern[0]]) != 0:                return False            else:                result = self.match(pattern[1:], str[len(d[pattern[0]]):], d, used)                if result:                    return True        else:            length = len(str) - (len(pattern) - 1)             for i in range(length):                cur = str[:i+1]                if cur in used and used[cur]:                    continue                used[cur] = True                d[pattern[0]] = cur                result = self.match(pattern[1:], str[len(cur):], d, used)                del d[pattern[0]]                if result:                    return True                del used[cur]                        return False

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