Python3解leetcode Isomorphic Strings
2020-12-13 05:57
标签：col with amp rmi for val ret cte add 问题描述： 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: Example 2: Example 3: 思路： 两个字符串对应位置上的字母要具有一一对应的关系，即假如s[1]=s,t[1]=t,则s对应的就是t,如果后面s[i]=s,则t[i]也必须为 t，若t[i]为其他字母，则判定为非同形态。 这种一一对应关系，首先考虑字典 代码： 如果字典中没有相应键值对，那么就在字典中加入该键值对；如果有该键值对，判断一下该元素是否和原存储的键值对相同，不相同则肯定是非同形态，如果相同则是到目前为止为同形态。 有一种特殊情况，就是不同的键，对应了相同的值，如s=‘ab‘,t=‘aa’，出现这种情况代表是非同形态的，通过判断键、值的个数来判断是否出现这种情况。 Python3解leetcode Isomorphic Strings 标签：col with amp rmi for val ret cte add 原文地址：https://www.cnblogs.com/xiaohua92/p/11158257.htmlInput: s =
"egg", t = "add"
Output: true
Input: s =
"foo", t = "bar"
Output: falseInput: s =
"paper", t = "title"
Output: true
Note:
You may assume both s and t have the same length.class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
table = dict()
for i in range(len(s)):
if s[i] not in table:
table[s[i]] = t[i]
elif table[s[i]] != t[i]:
return False
return len(table) == len(set(table.values()))
文章标题：Python3解leetcode Isomorphic Strings
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