【朱迪的LeetCode刷题笔记】205. Isomorphic Strings 同构字符串 #Easy #Python

Judie · 发表于 2023-5-23 12:51:59

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本帖最后由 Judie 于 2023-5-23 02:21 编辑

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.

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

Constraints:
1 <= s.length <= 5 * 104
t.length == s.length
s and t consist of any valid ascii character.

class Solution(object):
def isIsomorphic(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""

复制代码

class Solution {
public:
bool isIsomorphic(string s, string t) {
}
};

复制代码

Judy

class Solution(object):
def isIsomorphic(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
d = {}
for x, y in zip(s, t):
if x not in d:
if y not in d.values():
d[x] = y
else:
return False
elif d[x] != y:
return False
return True

复制代码

Gray
c++ bad version?

#include <iostream>
#include <map>
#include <algorithm>
class Solution {
public:
bool isIsomorphic(string s, string t) {
map<char,char> dict;
for(int i = 0; i < s.size(); i++){
vector<char> value;
for(auto it = dict.cbegin(); it != dict.cend(); ++it){
value.push_back(it->second);
}
if (dict.find(s[i])!=dict.end()){
s[i] = dict[s[i]];
}else {
cout << "cycle" << endl;
for(int j = 0; j < value.size();j++){
if(value[j] == t[i]){
return false;
}
}
dict[s[i]] = t[i];
s[i] = t[i];
}
}
if(s == t){
return true;
}else {
return false;
}
}
};

复制代码

c++ better version?

#include <iostream>
#include <map>
#include <algorithm>
class Solution {
public:
bool isIsomorphic(string s, string t) {
map<char,char> dict;
for(int i = 0; i < s.size(); i++){
vector<char> value;
if (dict.find(s[i])!=dict.end()){
s[i] = dict[s[i]];
}else {
for(auto it = dict.cbegin(); it != dict.cend(); ++it){
if(it->second == t[i]){
return false;
}
}
dict[s[i]] = t[i];
s[i] = t[i];
}
}
if(s == t){
return true;
}else {
return false;
}
}
};

复制代码

yinda_peng · 发表于 2023-5-23 14:31:16

有趣的同构

Judie · 发表于 2023-5-23 14:32:23

yinda_peng 发表于 2023-5-23 01:31
有趣的同构

谢谢！你也刷leetcode嘛！

yinda_peng · 发表于 2023-5-23 14:36:55

Judie 发表于 2023-5-23 14:32
谢谢！你也刷leetcode嘛！

是的，但只是偶尔刷，毕竟我需要花更多的时间在数学上，目前能够独立完成的水平仅限于一部分简单题，因为有些简单题的知识点没有接触过。主要使用C语言和Python

Judie · 发表于 2023-5-23 14:38:42

yinda_peng 发表于 2023-5-23 01:36
是的，但只是偶尔刷，毕竟我需要花更多的时间在数学上，目前能够独立完成的水平仅限于一部分简单题，因为 ...

了解！我打算从今天开始多刷刷

yinda_peng · 发表于 2023-5-23 14:39:54

Judie 发表于 2023-5-23 14:38
了解！我打算从今天开始多刷刷

加油！祝早日offer

Judie · 发表于 2023-5-23 14:40:38

yinda_peng 发表于 2023-5-23 01:39
加油！祝早日offer

谢谢！其实是为了学习算法准备考试

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