290. Word Pattern

zshouyi

于 2017-01-05 08:45:05 发布

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分类专栏： LeetCode Hash Table 文章标签： Hash Table Leetcode Word Pattern

本文链接：https://blog.youkuaiyun.com/zshouyi/article/details/54087903

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本文介绍了一种使用哈希映射解决字符串模式匹配问题的方法。该方法通过建立字符与非空单词之间的双射关系来判断输入的模式是否与给定字符串相匹配。文中提供了两种实现方案并详细解释了其工作原理。

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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.

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;
}