[刷题防痴呆] 0205 - 同构字符串 (Isomorphic

题目地址

https://leetcode.com/problems/isomorphic-strings/description/

题目描述

205. Isomorphic Strings

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.

思路

hash. 可以用数组, 也可以用hashmap. 从s走到t, 再从t映射到s.

关键点

注意, ASCII的范围是0-255. 所以用数组做hash的话, init数组长度为256.

代码

• 语言支持：Java

class Solution {
    public boolean isIsomorphic(String s, String t) {
        char[] sc = s.toCharArray();
        char[] tc = t.toCharArray();
        
        int[] map = new int[256];        
        for (int i = 0; i < sc.length; i++) {
            if (map[sc[i]] == 0) {
                map[sc[i]] = tc[i];
            } else {
                if (map[sc[i]] != tc[i]) {
                    return false;
                }
            }
        }
        
        int[] map2 = new int[256];
        for (int i = 0; i < tc.length; i++) {
            if (map2[tc[i]] == 0) {
                map2[tc[i]] = sc[i];
            } else {
                if (map2[tc[i]] != sc[i]) {
                    return false;
                }
            }
        }
        
        return true;
    }
}

// hashmap
class Solution {
    public boolean isIsomorphic(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }
        int n = s.length();
        Map<Character, Character> sMap = new HashMap<>();
        Map<Character, Character> tMap = new HashMap<>();

        for (int i = 0; i < n; i++) {
            char c1 = s.charAt(i);
            char c2 = t.charAt(i);
            if (sMap.containsKey(c1)) {
                if (sMap.get(c1) != c2) {
                    return false;
                }
            }
            if (tMap.containsKey(c2)) {
                if (tMap.get(c2) != c1) {
                    return false;
                }
            }

            sMap.put(c1, c2);
            tMap.put(c2, c1);
        }

        return true;
    }
}