leetcode205- Isomorphic Strings- easy

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.

用两个hashMap,分别表示s到t的字符映射和t到s的字符映射。如果出现了和以前的映射不匹配的情况,那就是不行的。

细节:1.两者长度不同的corner case 2.char的比较也要用equals不能用== 3.小心要两个hashmap,一个不够的比如"aa","ab"

实现:

class Solution {
    public boolean isIsomorphic(String s, String t) {
        
        if (s == null || t == null || s.length() != t.length()) {
            return false;
        }
        Map<Character, Character> mapA = new HashMap<>();
        Map<Character, Character> mapB = new HashMap<>();
        for (int i = 0; i < s.length(); i++) {
            Character sc = s.charAt(i);
            Character tc = t.charAt(i);
            if (mapA.containsKey(sc) && !mapA.get(sc).equals(tc) || mapB.containsKey(tc) && !mapB.get(tc).equals(sc)) {
                return false;
            }
            mapA.put(sc, tc);
            mapB.put(tc, sc);
        }
        return true;
    }
}
 
 
原文地址:https://www.cnblogs.com/jasminemzy/p/7838853.html