README_EN.md

205. Isomorphic Strings

中文文档

Description

Given two strings s and t, determine if they are isomorphic.

Two strings s and t 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

 

Constraints:

• 1 <= s.length <= 5 * 104
• t.length == s.length
• s and t consist of any valid ascii character.

Solutions

Python3

class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        a2b, b2a = {}, {}
        n = len(s)
        for i in range(n):
            a, b = s[i], t[i]
            if (a in a2b and a2b[a] != b) or (b in b2a and b2a[b] != a):
                return False
            a2b[a] = b
            b2a[b] = a
        return True

class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        m1, m2 = [0] * 256, [0] * 256
        for i in range(len(s)):
            c1, c2 = ord(s[i]), ord(t[i])
            if m1[c1] != m2[c2]:
                return False
            m1[c1] = m2[c2] = i + 1
        return True

Java

class Solution {
    public boolean isIsomorphic(String s, String t) {
        int n = s.length();
        Map<Character, Character> a2b = new HashMap<>();
        Map<Character, Character> b2a = new HashMap<>();
        for (int i = 0; i < n; ++i) {
            char a = s.charAt(i), b = t.charAt(i);
            if ((a2b.containsKey(a) && a2b.get(a) != b) || (b2a.containsKey(b) && b2a.get(b) != a)) return false;
            a2b.put(a, b);
            b2a.put(b, a);
        }
        return true;
    }
}

class Solution {
    public boolean isIsomorphic(String s, String t) {
        int[] m1 = new int[256];
        int[] m2 = new int[256];
        for (int i = 0; i < s.length(); ++i) {
            char c1 = s.charAt(i);
            char c2 = t.charAt(i);
            if (m1[c1] != m2[c2]) {
                return false;
            }
            m1[c1] = i + 1;
            m2[c2] = i + 1;
        }
        return true;
    }
}

C++

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        vector<int> m1(256);
        vector<int> m2(256);
        for (int i = 0; i < s.size(); ++i)
        {
            if (m1[s[i]] != m2[t[i]]) return 0;
            m1[s[i]] = i + 1;
            m2[t[i]] = i + 1;
        }
        return 1;
    }
};

Go

func isIsomorphic(s string, t string) bool {
	m1, m2 := make([]int, 256), make([]int, 256)
	for i := 0; i < len(s); i++ {
		if m1[s[i]] != m2[t[i]] {
			return false
		}
		m1[s[i]] = i + 1
		m2[t[i]] = i + 1
	}
	return true
}

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