【VIP】205. Isomorphic Strings【E】【61】【太经典了】

本文讨论如何通过哈希表判断两个字符串是否同构，详细解释了实现思路并提供了代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Subscribe to see which companies asked this question

判断是否同构

最开始想的是，每个字母用数字替换，可是后来意识到，字符串里可能包含数字，不行

然后，用，每个字符得逐个删除，结果超时了

看看答案，人家用的是哈希表，首先一共就几百个字符

这儿有两个关键，一个是，什么时候退出，也就是return False，那就是，如果如下面代码的条件，为什么呢？因为如果只出现一次的话，不需要判定是不是相等，因为直接同构了

第二个关键，那就是，i+1，否则的话，跟0相等了，这样会出错

class Solution(object):
    def isIsomorphic(self, s, t):
        res1 = [0] * 512#len(s)
        res2 = [0] * 512#len(s)

        for i in xrange(len(s)):
            if res1[ord(s[i])] != res2[ord(t[i])]:
                return False
            res1[ord(s[i])] = i + 1
            res2[ord(t[i])] = i + 1

        return True