Isomorphic Strings (E)

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.

题意

判断两个字符串是否为“同构”字符串，即长度相同，且s与t的每个位置上的字符都可以一一映射。

思路

最直接的方法就是开两个HashMap，将映射关系全部记录下来再进行判断，但代价就是用时太长；

如果将每个字符串中的每一个字符都映射成它在该字符串中第一次出现的位置下标，那么根据同构字符串的性质，得到的两个数组应该是完全相同的，这样就不需要用到HashMap，例如下：

代码实现 - Hash

class Solution {
    public boolean isIsomorphic(String s, String t) {
        HashMap<Character, Character> st = new HashMap<>();
        HashMap<Character, Character> ts = new HashMap<>();
        for (int i = 0; i < s.length(); i++) {
            char a = s.charAt(i);
            char b = t.charAt(i);
            if (st.containsKey(a) && st.get(a) != b || ts.containsKey(b) && ts.get(b) != a) {
                return false;
            } else {
                st.put(a, b);
                ts.put(b, a);
            }
        }
        return true;
    }
}

代码实现 - 无Hash

class Solution {
    public boolean isIsomorphic(String s, String t) {
        for (int i = 0; i < s.length(); i++) {
            if (s.indexOf(s.charAt(i)) != t.indexOf(t.charAt(i))) {
                return false;
            }
        }
        return true;
    }
}