205. Isomorphic Strings java solutions

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.

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.

Subscribe to see which companies asked this question

 

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s==null || t == null || s.length() != t.length()) return false;
        HashMap<Character, Character> hms = new HashMap<Character, Character>();
        HashMap<Character, Character> hmt = new HashMap<Character, Character>();
        int[] tmp = new int[s.length()];
        for(int i = 0; i< s.length();i++){
            Character ss = s.charAt(i);
            Character tt = t.charAt(i);
            if(hms.containsKey(ss) || hmt.containsKey(tt)){
                if(hms.get(ss) != tt || hmt.get(tt) != ss) return false;
            }else{
                hms.put(ss,tt);
                hmt.put(tt,ss);
            }
        }
        return true;
    }
}

证明两个字符串是同构的，同个下标的a->b， b->a，即可。

做法二：

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        int[] times1 = new int[128], times2 = new int[128];
        int len = s.length();
        for (int i = 0, smax = 0, tmax = 0; i < len; ++i) {
            char sc = s.charAt(i);
            char tc = t.charAt(i);
            if (times1[sc] == 0) {
                times1[sc] = ++smax;
            }
            if (times2[tc] == 0) {
                times2[tc] = ++tmax;
            }
            if (times1[sc] != times2[tc])
                return false;
        }
        return true;
    }
}

http://blankj.com/404.html