leetcode 290. Word Pattern

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:

1. pattern = "abba", str = "dog cat cat dog" should return true.
2. pattern = "abba", str = "dog cat cat fish" should return false.
3. pattern = "aaaa", str = "dog cat cat dog" should return false.
4. 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;    }}