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:
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.
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:
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()))
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。