Isomorphic Strings [LeetCode]

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.

题意：对于两个字符串S和T分别将S中的字符对应到T中的字符，和从T中的字符映射的S中的字符

最开始理解为只有从S中的字符映射到T，贡献一个WA，后来改为每个字符都只能对应一个字符，同样WA了。最后才发现需要从S中映射到T,从T中映射到S，才Accept。实现比较简单可以直接看代码；

PS：自己需要注意的地方map.find()函数如果没有找到返回的不是false,而是map.end();

实现代码(c++):

#include<cstdio>#include<cstring>#include<string>#include<cstdlib>#include<map>#include<iostream>using namespace std;class Solution {public:    bool isIsomorphic(string s, string t) {        if(s.size()!= t.size())        {        return false;        }        map<char,char> mapS;        map<char,char> mapT;        for(int i = 0;i<s.size();i++)        {        if(mapS.find(s[i])!=mapS.end())        {        //cout << s[i] << " " << m[s[i]] << endl;        if(mapS[s[i]] != t[i])        {        return false;        }        }        else{        mapS[s[i]] = t[i];}if(mapT.find(t[i])!=mapT.end())        {        //cout << s[i] << " " << m[s[i]] << endl;        if(mapT[t[i]] != s[i])        {        return false;        }        }        else{        mapT[t[i]] = s[i];}        }        return true;    }};int main(){string s1,s2;Solution s;while(cin >> s1 >> s2){if(s.isIsomorphic(s1,s2)){cout << "yes" << endl;}else{ cout << "no" << endl;}}return 0;}