205. Isomorphic Strings

205. Isomorphic Strings

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

Two strings s and t 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

Constraints:

• 1 <= s.length <= 5 * 104

• t.length == s.length

• s and t consist of any valid ascii character.

My Solutions:

方法1：

建立sm和tm两个map，对s和t里的每一个字母，必须在另一个map里相互对应。比如sm[e] = a, tm[a] = e.

方法2：

map里不储存字母，而是储存出现的index。s和t里出现的对应字母应该index相同。

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        
        // char[] s = sString.toCharArray();
        // char[] t = tString.toCharArray();

        // int length = s.length;
        // if(length != t.length) return false;

        // char[] sm = new char[256];
        // char[] tm = new char[256];

        // for(int i=0; i<length; i++){
        //     char sc = s[i];
        //     char tc = t[i];
        //     if(sm[sc] == 0 && tm[tc] == 0){
        //         sm[sc] = tc;
        //         tm[tc] = sc;
        //     }else{
        //         if(sm[sc] != tc || tm[tc] != sc){
        //             return false;
        //         }
        //     }
        // }
        // return true;
        
        
        int[]  mapS= new int[256], mapT = new int[256];
        int len = s.length();
        for (int i = 0; i < len; i++) {
            if (mapS[s.charAt(i)] != mapT[t.charAt(i)]) return false;
            mapS[s.charAt(i)] = i + 1;
            mapT[t.charAt(i)] = i + 1;
        }
        return true;
    }
}