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.

方法一

public class Solution {/* * 算法思想： * 将pattern中出现的字符与str中出现的字符串，根据顺序当作匹配关系，然后存放入map中，然后依次进行比较 */public boolean wordPattern(String pattern, String str) {if (pattern == null || str == null) {return false;}char[] charArr = pattern.toCharArray(); // 将pattern转化为字符数组String[] stringArr = str.split("\\s+"); // 按空格进行分隔if (charArr.length != stringArr.length) {return false;}// 用于存放patter中字符与String的对应关系Map<Character, String> map = new HashMap<Character, String>();for (int i = 0; i < charArr.length; i++) {if (map.containsKey(charArr[i])) {if (!map.get(charArr[i]).equals(stringArr[i])) {return false;}}// 保证key和value都唯一Set<Character> keySet = map.keySet(); // 得到key的集合keySet.remove(charArr[i]);for (Character chr : keySet) {if (map.get(chr).equals(stringArr[i])) {return false;}}map.put(charArr[i], stringArr[i]);}return true;}}

方法二

public class Solution {public boolean wordPattern(String pattern, String str) {String[] words = str.split("\\s+");    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;}}