Isomorphic Strings 判断俩等长字符串对应位置出现的字母是否类比相等

Given two strings s and t, determine if they are isomorphic.

Two strings are isomorphic if the characters in s can be replaced to gett.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.

For example,
Given "egg", "add", return true.

Given "foo", "bar", return false.

Given "paper", "title", return true.

Note:
You may assume both s and t have the same length.

 
public class Solution {
    public boolean isIsomorphic(String s, String t) {
        HashMapmaps=new HashMap();
        HashMapmapt=new HashMap();
         
        for(Integer i=0;i            if(maps.put(s.charAt(i),i)!=mapt.put(t.charAt(i),i)) return false; //put返回的key锁对应的上一个值
        }
        return true;
    }
}

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        HashMapmaps=new HashMap();
        HashSetset=new HashSet();//set是一元的并且不允许有重复元素
         
        for(Integer i=0;i            if(maps.containsKey(s.charAt(i))){
               if (maps.get(s.charAt(i))!=t.charAt(i)) return false;
           }
           else{
               if(set.contains(t.charAt(i)))    return false;//注意这儿要先行判断
               else{
                   maps.put(s.charAt(i),t.charAt(i));//key，value为字符串s和t里边对应位置的字母
                   set.add(t.charAt(i));//set维护t里边的字母                
                    
               }
           }
        }
        return true;
    }
}

/*public class Solution {
    public boolean isIsomorphic(String s, String t) {
        HashMapmaps=new HashMap();
        HashMapmapt=new HashMap();
         
        for(Integer i=0;i            if(maps.containsKey(s.charAt(i))) maps.put(s.charAt(i),maps.get(s.charAt(i))+1);
           else maps.put(s.charAt(i),1);
        }
        for(Integer i=0;i            if(mapt.containsKey(t.charAt(i))) mapt.put(t.charAt(i),mapt.get(t.charAt(i))+1);
           else mapt.put(t.charAt(i),1);
        }
        for(Integer i=0;i             if(maps.get(s.charAt(i))!= mapt.get(t.charAt(i)))  return false; //有错，把hashmap当做有序存放的了，记得他只跟key有关
        }
        return true;
    }
}*/