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# Word Pattern

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** word in `str`.

**Example 1:**

```
Input: pattern = "abba", str = "dog cat cat dog"
Output: true
```

**Example 2:**

```
Input:pattern = "abba", str = "dog cat cat fish"
Output: false
```

**Example 3:**

```
Input: pattern = "aaaa", str = "dog cat cat dog"
Output: false
```

**Example 4:**

```
Input: pattern = "abba", str = "dog dog dog dog"
Output: false
```

**Notes:**\
You may assume `pattern` contains only lowercase letters, and `str`contains lowercase letters that may be separated by a single space.

分析

2个map，存的是当前Index，每次比较map值，也就是之前index。

注意get（key,defaultvalue）用法，defaultvalue要用-1 不能用0.

```
class Solution:
    def wordPattern(self, pattern: str, str: str) -> bool:
        str = str.split(' ')
        if len(str) != len(pattern):
            return False
        # wordsMap = {}
        # for i in range(len(str)):
        #     if pattern[i] in wordsMap and wordsMap[pattern[i]] != str[i]:
        #          return False
        #     if pattern[i] not in wordsMap:
        #         if str[i] in wordsMap.values():
        #             return False
        #         wordsMap[pattern[i]] = str[i]
        # return True
        s,p = {},{}
        for i in range(len(str)):
            if s.get(str[i],-1) != p.get(pattern[i],-1): #default index必须是-1,因为0是有效Index
                return False
            s[str[i]] = p[pattern[i]] = i
        return True
            
        
```