题目:
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.
思路:
- 思路1: 为字符串s和t设置两个map,使得对应字符形成一一对应的关系。
- 思路2: 对于字符串s和t,设置两个map,保存上次出现的位置。每次迭代时,判断两字符(s[i]和t[i])上次出现的位置是否相同。
代码实现:
思路1:
class Solution {
public:
bool isIsomorphic(string s, string t) {
unordered_map<char, char> s_map;
unordered_map<char, char> t_map;
for (int i = 0; i < s.size(); ++i){
if (s_map.find(s[i]) != s_map.end()){
if (s_map[s[i]] != t[i]){
return false;
}
}else{
if (t_map.find(t[i]) != t_map.end()){
return false;
}
s_map[s[i]] = t[i];
t_map[t[i]] = s[i];
}
}
return true;
}
};
思路2:
class Solution {
public:
bool isIsomorphic(string s, string t) {
unordered_map<char, int> s_map;
unordered_map<char, int> t_map;
for (int i = 0; i < s.size(); ++i){
if (s_map.find(s[i]) != s_map.end() && t_map.find(t[i]) != t_map.end()){
if (s_map[s[i]] != t_map[t[i]]){
return false;
}
}else if(s_map.find(s[i]) == s_map.end() && t_map.find(t[i]) == t_map.end()){
s_map[s[i]] = i;
t_map[t[i]] = i;
}else{
return false;
}
}
return true;
}
};