290. Word Pattern

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本文介绍了一种检查字符串是否遵循给定模式的方法。通过两个不同的实现方案，确保字符串中的单词与模式中的字符一一对应。

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.

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;
	}
}