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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

Explanation:

The strings s and t can be made identical by:

  • Mapping 'e' to 'a'.
  • Mapping 'g' to 'd'.

Example 2:

Input: s = "foo", t = "bar"

Output: false

Explanation:

The strings s and t can not be made identical as 'o' needs to be mapped to both 'a' and 'r'.

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

Solution 1: Hash Table or Array

We can use two hash tables or arrays $d_1$ and $d_2$ to record the character mapping relationship between $s$ and $t$.

Traverse $s$ and $t$, if the corresponding character mapping relationships in $d_1$ and $d_2$ are different, return false, otherwise update the corresponding character mapping relationships in $d_1$ and $d_2$. After the traversal is complete, it means that $s$ and $t$ are isomorphic, and return true.

The time complexity is $O(n)$ and the space complexity is $O(C)$. Where $n$ is the length of the string $s$; and $C$ is the size of the character set, which is $C = 256$ in this problem.

Python3

class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        d1 = {}
        d2 = {}
        for a, b in zip(s, t):
            if (a in d1 and d1[a] != b) or (b in d2 and d2[b] != a):
                return False
            d1[a] = b
            d2[b] = a
        return True

Java

class Solution {
    public boolean isIsomorphic(String s, String t) {
        Map<Character, Character> d1 = new HashMap<>();
        Map<Character, Character> d2 = new HashMap<>();
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            char a = s.charAt(i), b = t.charAt(i);
            if (d1.containsKey(a) && d1.get(a) != b) {
                return false;
            }
            if (d2.containsKey(b) && d2.get(b) != a) {
                return false;
            }
            d1.put(a, b);
            d2.put(b, a);
        }
        return true;
    }
}

C++

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        int d1[256]{};
        int d2[256]{};
        int n = s.size();
        for (int i = 0; i < n; ++i) {
            char a = s[i], b = t[i];
            if (d1[a] != d2[b]) {
                return false;
            }
            d1[a] = d2[b] = i + 1;
        }
        return true;
    }
};

Go

func isIsomorphic(s string, t string) bool {
	d1 := [256]int{}
	d2 := [256]int{}
	for i := range s {
		if d1[s[i]] != d2[t[i]] {
			return false
		}
		d1[s[i]] = i + 1
		d2[t[i]] = i + 1
	}
	return true
}

TypeScript

function isIsomorphic(s: string, t: string): boolean {
    const d1: number[] = new Array(256).fill(0);
    const d2: number[] = new Array(256).fill(0);
    for (let i = 0; i < s.length; ++i) {
        const a = s.charCodeAt(i);
        const b = t.charCodeAt(i);
        if (d1[a] !== d2[b]) {
            return false;
        }
        d1[a] = i + 1;
        d2[b] = i + 1;
    }
    return true;
}

Rust

use std::collections::HashMap;
impl Solution {
    fn help(s: &[u8], t: &[u8]) -> bool {
        let mut map = HashMap::new();
        for i in 0..s.len() {
            if map.contains_key(&s[i]) {
                if map.get(&s[i]).unwrap() != &t[i] {
                    return false;
                }
            } else {
                map.insert(s[i], t[i]);
            }
        }
        true
    }

    pub fn is_isomorphic(s: String, t: String) -> bool {
        let (s, t) = (s.as_bytes(), t.as_bytes());
        Self::help(s, t) && Self::help(t, s)
    }
}

C#

public class Solution {
    public bool IsIsomorphic(string s, string t) {
        int[] d1 = new int[256];
        int[] d2 = new int[256];
        for (int i = 0; i < s.Length; ++i) {
            var a = s[i];
            var b = t[i];
            if (d1[a] != d2[b]) {
                return false;
            }
            d1[a] = i + 1;
            d2[b] = i + 1;
        }
        return true;
    }
}

Solution 2

Python3

class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        d1, d2 = [0] * 256, [0] * 256
        for i, (a, b) in enumerate(zip(s, t), 1):
            a, b = ord(a), ord(b)
            if d1[a] != d2[b]:
                return False
            d1[a] = d2[b] = i
        return True

Java

class Solution {
    public boolean isIsomorphic(String s, String t) {
        int[] d1 = new int[256];
        int[] d2 = new int[256];
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            char a = s.charAt(i), b = t.charAt(i);
            if (d1[a] != d2[b]) {
                return false;
            }
            d1[a] = i + 1;
            d2[b] = i + 1;
        }
        return true;
    }
}