205. Isomorphic Strings [easy] Python3 C++

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最新推荐文章于 2025-05-14 17:02:27 发布

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本文介绍了一种用于判断两个字符串是否为同构字符串的有效算法。通过使用哈希映射的方法，确保了所有出现的字符能够被一一对应，从而验证了两个字符串的模式是否一致。文章提供了Python和C++两种实现方式，并附带详细的代码解析。

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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.

题目要求：判断两个字符串的模式相同，及用一个字符串中的字符替换相同位置的第二个字符串中的字符，得到的字符串与第一个字符串相同。

Python3

class Solution:
    def isIsomorphic(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        hashmap = {}
        for i in range(len(s)):
            if s[i] not in hashmap:
                hashmap[s[i]] = t[i]
            elif hashmap[s[i]] != t[i]:
                return False
        
        val = [hashmap[i] for i in hashmap]
        return len(val) == len(set(val))

C++(AC里看到的，比较巧妙)

class Solution {
public:
    bool isIsomorphic(string s, string t) {
        vector<int> v1(256, -1), v2(256, -1);
        for(int i=0; i<s.size(); ++i)
        {
            if(v1[s[i]] != v2[t[i]])
                return false;
            
            v1[s[i]] = v2[t[i]] = i;
        }
        
        return true;
    }
};