*205. Isomorphic Strings (string & map idea)

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.

 

Accepted solution:

create an array: map all the character and count the value as the index of each characters in string. start from 1

e.g.: egg & dda & add

a['e'] = 1 a['g'] = 2 -> 3 ; a['d'] = 1 -> 2  a['a'] 3 ; a['a'] = 1   a['d'] 2 -> 3

class Solution {
    public boolean isIsomorphic(String s, String t) {
        //letter , index of string
        //e.g. egg : e: 1 g:2 g:3
        if(s.length()!=t.length()) return false;
        int[] a = new int[256];
        int[] b = new int[256];
        for(int i = 0; i<s.length(); i++){
            if(a[s.charAt(i)] != b[t.charAt(i)]) return false;
            //update the character 
            a[s.charAt(i)] = i+1; // why plus 1 for the case aa & ab
            b[t.charAt(i)] = i+1;
        }
        return true;
    }
}

Good idea make the code clean and easy. Think before coding.

转载于:https://www.cnblogs.com/stiles/p/leetcode205.html