leetcode with python:205. Isomorphic Strings

题目:

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.

给定两个字串,判断一字串中的字元是否可透过字元替换而变成另一个字串
要特别注意的是,替换的关係仅能是一对一映射的
ex:(1)input:"egg","add"=>output: true(e换a,g换d)
(2)input:"foo","bar"=>output: false(f换b,o换a,o换r,很明显o的替换关係不是1对1映射)

我们透过建立hash map,纪录对应关係,观察是否有非一对一关係

class Solution:    def isIsomorphic(self, s: str, t: str) -> bool:        d={}        for i in range(len(s)):            if s[i] not in d:                 if t[i] not in d.values():                    d[s[i]]=t[i]                else:                    return False            else:                if d[s[i]]!=t[i]:                    return False                return True

s,t个别表两字串,然后我们设dictionary为d存放对应关係
从头开始存放关係(s[i]是key,t[i]是value),会出现以下几种可能

(1)s[i]不存在key中且t[i]不存在value中:发现新对应关係,纪录

(2)s[i]不存在key中但t[i]存在value中:发现非一对一关係,return False

(3)s[i]存在key中且对应值和t[i]相同:发现重複的关係而已,无视

(4)s[i]存在key中但对应值和t[i]不同:发现非一对一关係,return False

若纪录完都没发现非一对一关係就return True
最后执行时间30ms(faster than 99.74%)

那我们下题见