leetcode 290. Word Pattern

最新推荐文章于 2024-01-10 12:35:36 发布

原创最新推荐文章于 2024-01-10 12:35:36 发布 · 380 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode 题目

OJ & Leetcode 专栏收录该内容

46 篇文章

订阅专栏

本文介绍了一种基于双哈希表的模式匹配算法，用于判断给定的字符串是否遵循指定的模式。通过实例演示了如何实现全匹配，并考虑了各种边界情况。

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.

给你一个匹配模式，如果给定的字符串符合这个模式，则为true，否则为false

利用两个哈希表完成,需要考虑到一些边界条件。

两个哈希表，分别存放左边的字符和右边的字符串，以及右边的字符串和左边的字符，判断匹配的时候，两个都查询一下。

import java.util.*;
class Solution {
    public boolean wordPattern(String pattern, String str) {
         if(!str.contains(" ")){
        if(pattern.equals(str)){
            if(str.length()==1&&pattern.length()==1)
            return true;
            else 
            return false;
        }else{
            if(pattern.length()==1)
                return true;
            return false;
        }
         }
        
      String [] strs = str.split(" ");
        if(strs.length!=pattern.length()){
            return false;
        }
        Map<String,String> map = new HashMap<String,String>();
        Map<String,String> map2 = new HashMap<String,String>();
        for(int i=0;i<pattern.length();i++){
              String ss = pattern.charAt(i)+"" ;
              String aa = map.get(ss);
              if(map.containsKey(ss)&&!aa.equals(strs[i])){
                  return false;
              }else if(!map.containsKey(pattern.charAt(i)+"")&&!map2.containsKey(strs[i])){
                  map.put(ss,strs[i]);
                  map2.put(strs[i],ss);
              }else if(!map.containsKey(ss)&&map2.containsKey(strs[i])){
                  return false;
              }
              
        }
        return true;
    }
}