290. Word Pattern_give a pattern and csdn-优快云博客

本文介绍了一种检查字符串是否遵循特定模式的算法。通过两个示例解决方案展示了如何使用哈希表和映射来确保模式与字符串完全匹配，即模式中的每个字符与字符串中的非空单词之间存在一一对应关系。

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.

解法一：

import java.util.Hashtable;
public class Solution {
    char index = 'a';
    public boolean wordPattern(String pattern, String str) {
        if (pattern.length() != str.split(" ").length)
			return false;
		Map<String, Character> table = new Hashtable<String, Character>();
		StringBuilder sb = new StringBuilder();
		for(int i=0;i<pattern.length();i++){
			if(table.get(String.valueOf(pattern.charAt(i)))==null){
				table.put(String.valueOf(pattern.charAt(i)), index++);
			}
			sb.append(table.get(String.valueOf(pattern.charAt(i))));
		}
		String[] sub = str.split(" ");
		index = 'a';
		table.clear();
		for (int i = 0; i < sub.length; i++) {
			if (table.get(sub[i]) == null)
				table.put(sub[i], index++);
			if (table.get(sub[i]) != sb.charAt(i))
				return false;
		}
		return true;
    }
}

解法二：

public class Solution {
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;
	}
}

注：HashMap中put（Key k，Value v）方法的返回值：若键为k的键值对不存在，则返回null；否则返回put执行之前map中键为k的键值对的Value值。