290. 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.

Examples:

1. pattern = "abba", str = "dog cat cat dog" should return true.
2. pattern = "abba", str = "dog cat cat fish" should return false.
3. pattern = "aaaa", str = "dog cat cat dog" should return false.
4. 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.

Credits:
Special thanks to @minglotus6 for adding this problem and creating all test cases.

这里注意不是比较worsd的首字母，是针对整个word。hashmap求解，代码如下：

public class Solution {    public boolean wordPattern(String pattern, String str) {        HashMap<Character, String> hs = new HashMap<Character, String>();        String[] strs = str.trim().split("\\s");        if (pattern.length() != strs.length) {            return false;        }        for (int i = 0; i < strs.length; i ++) {            if(hs.containsKey(pattern.charAt(i))) {                if (!strs[i].equals(hs.get(pattern.charAt(i)))) {                    return false;                }            } else {                if (hs.containsValue(strs[i])){                    return false;                }                hs.put(pattern.charAt(i), strs[i]);            }        }        return true;    }}

另一种巧妙的方法是利用hashmap 的put 方法，put 会返回key 之前的value 或者 null（无key 时）。代码如下：

public boolean wordPattern(String pattern, String str) {    String[] words = str.split(" ");    if (words.length != pattern.length())        return false;    Map index = new HashMap();    for (Integer i=0; i<words.length; ++i)        if (index.put(pattern.charAt(i), i) != index.put(words[i], i))            return false;    return true;}