[leetcode]205. Isomorphic Strings 同构字符串

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

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

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.

Example 1:

Input: s = "egg", t = "add"
Output: true

Example 2:

Input: s = "foo", t = "bar"
Output: false

Example 3:

Input: s = "paper", t = "title"
Output: true

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

思路1（用两个map）

1.  scan char a from S and char b from T in the same time

2. use two int array to mimic hash table: mapAB, mapBA 

3. if mapAB[a] == 0 , means I didn't mapping it,  assign current b to mapAB[a]

    otherwise,   means I did mapping it,  check  b == mapAB[a] ?

4. do the same operation for mapBA

5. why checking two maps at the same time?  Coz S: egg -> T: aaa   return true

   then we still check T: aaa ->  S: egg

代码

class Solution {
    public boolean isIsomorphic(String s, String t) { 
        if(s.length()!=t.length()) return false;   
        // uses the arrays to mimic two hash table
         int [] mapAB = new int[256];//ASCII characters
         int [] mapBA = new int[256];
         for(int i = 0; i< s.length();i++){
            char a = s.charAt(i);
            char b = t.charAt(i);  
            // mapAB[a]!=0 means I already mapping it
            if(mapAB[a]!=0){
                if(mapAB[a]!=b) return false;
            }
            // mapAB[a]==0 means I haven't mapping it 
            else{
                mapAB[a]= b;
            }
            
            // why checking two map? coz S:egg T: aaa would return true if only checking mapAB
            if(mapBA[b]!=0){
                if(mapBA[b]!=a) return false;
            }else{
                mapBA[b] = a;
            }           
        } 
        return true;   
    }
}

思路2（只用一个map）

1. scan either S or T(assuming they have same length)
2. store the idx of current char in both strings.
    if previously stored idx are different from current idx, return false

举例：

S: egg      T: aaa
m[e] = 1    m[a+256] = 1
m[g] = 2   occurs that previous m[a+256] = 1 return false

代码

class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s.length() != t.length()) return false;
        int[] m = new int[512];
        for (int i = 0; i < s.length(); i++) {
            if (m[s.charAt(i)] != m[t.charAt(i)+256]) return false;
            m[s.charAt(i)] = m[t.charAt(i)+256] = i+1;
        }
        return true;
    }
}

/* Why not m[s.charAt(i)] = m[t.charAt(i)+256] = i ? 
            coz 0 is the default value, we should not use it. otherwise we cannot distinguish between 
            the default maker and the the marker we made.
            S: aa   T: ab
            i= 0： m[a] = 0  m[a+256] = 0
                             but m[b+256] is defaulted 0 

*/

followup1:

如果输入K个string，判断其中至少两个是Isomorphic Strings， 返回boolean

思路

1. 将所有的given string都转成同一种pattern

ex. foo -> abb
ex. gjk -> abc
ex. pkk -> abb

2. 用一个hashmap来存 transfered word 和其出现的频率。

               key  ： value(frequency)

ex. foo -> abb ： 1 
ex. gjk ->  abc :  1 
ex. pkk -> abb :  1+1   return true  

代码

   public boolean findIsomorphic(String[] input) {
        // key: transWord,  value: its corresponding frequency
        Map<String, Integer> map = new HashMap<>();
        for (String s : input) {
            // transfer each String into same pattern
            String transWord = transfer(s);
            if (!map.containsKey(transWord)) {
                map.put(transWord, 1);
            }
            // such transWord pattern already in String[]
            else {
                return true;
            }
        }
        return false;
    }

    /* pattern:  every word start with 'a'
       when comes a new letter, map it to cur char, 
       and increase the value of cur cha
     */
    private String transfer(String word) {
        Map<Character, Character> map = new HashMap<>();
        StringBuilder sb = new StringBuilder();
        char cur = 'a';
        for (char letter : word.toCharArray()) {
            if (!map.containsKey(letter)) {
                map.put(letter, cur);
                cur++;
            }
            sb.append(map.get(letter));
        }
        return sb.toString();
    }

followup2:

如果输入K个string， 判断其中任意两两是Isomorphic Strings，返回boolean

即给定K个string都能化成同一种等值的pattern