# 290. Word Pattern 单词规律

@TOC

## # 题目描述

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.

Examples:

``````pattern = "abba", str = "dog cat cat dog" should return true.
pattern = "abba", str = "dog cat cat fish" should return false.
pattern = "aaaa", str = "dog cat cat dog" should return false.
pattern = "abba", str = "dog dog dog dog" should return false.
``````

Notes:

• You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.

## # 解题方法

``````public class Solution {
public boolean wordPattern(String pattern, String str) {
HashMap<Character, String> map = new HashMap<Character, String>();
String[] words = str.split(" ");
if(words.length != pattern.length()){
return false;
}
for(int i = 0; i < words.length; i++){
String word = words[i];
char temp = pattern.charAt(i);
if(map.containsKey(temp)){
if(!map.get(temp).equals(word)){
return false;
}
}else{
if(map.containsValue(word)){
return false;
}
map.put(temp, word);
}
}
return true;
}
}
``````

``````class Solution(object):
def wordPattern(self, pattern, str):
"""
:type pattern: str
:type str: str
:rtype: bool
"""
strs = str.split()
if len(pattern) != len(strs):
return False
d = dict()
for i, p in enumerate(pattern):
if p not in d:
d[p] = strs[i]
else:
if d[p] != strs[i]:
return False
d = dict()
for i, p in enumerate(strs):
if p not in d:
d[p] = pattern[i]
else:
if d[p] != pattern[i]:
return False
return True
``````

## # 日期

2017 年 5 月 19 日 2018 年 11 月 24 日 —— 周六快乐