290. Word Pattern_word upattern-优快云博客

本文介绍了一种检查字符串是否遵循给定模式的算法实现。通过建立字符与单词间的映射关系，确保模式与字符串中各部分一一对应。文章提供了Java代码示例，并探讨了该问题与等同字符串问题之间的联系。

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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.

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

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Solution:

Tips:

seems to 205. Isomorphic Strings

Java Code:

public class Solution {
    public boolean wordPattern(String pattern, String str) {
        char[] p = pattern.toCharArray();
        String[] s = str.split(" ");
        if (p == null && s == null) {
            return true;
        }
        if (p == null && s != null || p != null && s == null || p.length != s.length){
            return false;
        }
        
        int[] pTable = new int[256];
        Map<String, Integer> sTable = new HashMap<>();
        for (int i = 0; i < s.length; i++) {
            sTable.put(s[i], 0);
        }
        
        for (int i = 0; i < p.length; i++) {
            if (!sTable.containsKey(s[i]) || (pTable[p[i]] != sTable.get(s[i]))) {
                return false;
            }
            pTable[p[i]] = i + 1;
            sTable.put(s[i], i + 1);
        }
        return true;
    }
}