Leetcode: 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.

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.

Best: O(N) time, constant space

public class Solution {
    public boolean isIsomorphic(String s1, String s2) {
        int[] m = new int[256];
        int[] n = new int[256];
        for (int i = 0; i < s1.length(); i++) {
            if (m[s1.charAt(i)] != n[s2.charAt(i)]) return false;
            m[s1.charAt(i)] = n[s2.charAt(i)] = i+1;
        }
        return true;
    }
}

复杂度

时间 O(N) 空间 O(N)

思路

用一个哈希表记录字符串s中字母到字符串t中字母的映射关系，一个集合记录已经映射过的字母。或者用两个哈希表记录双向的映射关系。这里不能只用一个哈希表，因为要排除egg->ddd这种多对一的映射。

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        Map<Character, Character> map = new HashMap<Character, Character>();
        Set<Character> set = new HashSet<Character>();
        if(s.length() != t.length()) return false;
        for(int i = 0; i < s.length(); i++){
            char sc = s.charAt(i), tc = t.charAt(i);
            if(map.containsKey(sc)){
                // 如果已经给s中的字符建立了映射，检查该映射是否和当前t中字符相同
                if(tc != map.get(sc)) return false;
            } else {
                // 如果已经给t中的字符建立了映射，就返回假，因为出现了多对一映射
                if(set.contains(tc)) return false;
                map.put(sc, tc);
                set.add(tc);
            }
        }
        return true;
    }
}