算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## Word Pattern


### 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**:

1. pattern = `"abba"`, str = `"dog cat cat dog"` should return true.
1. pattern = `"abba"`, str = `"dog cat cat fish"` should return false.
1. pattern = `"aaaa"`, str = `"dog cat cat dog"` should return false.
1. 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.


### 分析

本题跟 "Isomorphic Strings" 很类似,用两个HashMap, 记录从字符到字符串和字符串到字符的映射。


### 代码

{% if book.java %}
```java
// Word Pattern
// Time Complexity: O(n), Space Complexity: O(n)
public class Solution {
public boolean wordPattern(String pattern, String str) {
String[] words = str.split(" ");
if (words.length != pattern.length()) return false;

final Map map1 = new HashMap<>();
final Map map2 = new HashMap<>();
for (int i = 0; i < words.length; ++i) {
final Character key1 = pattern.charAt(i);
if (map1.containsKey(key1)) {
final String value = map1.get(key1);
if (!value.equals(words[i])) return false;
} else {
map1.put(key1, words[i]);
}

final String key2 = words[i];
if (map2.containsKey(key2)) {
final char value = map2.get(key2);
if (value != pattern.charAt(i)) return false;
} else {
map2.put(key2, pattern.charAt(i));
}
}
return true;
}
}
```
{% endif %}


### 相关题目

* [Isomorphic Strings](isomorphic-strings.md)