[Leetcode]205. Isomorphic Strings
2016-07-13 19:30
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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
return true.
Given
return false.
Given
return true.
Note:
You may assume both s and t have the same length.
class Solution {
public:
bool isIsomorphic(string s, string t) {
map<char, char> m;
for (int i = 0; i != s.size(); ++i) {
if (m.find(s[i]) == m.end())
m[s[i]] = t[i];
else if (m[s[i]] != t[i])
return false;
}
m.clear();
for (int i = 0; i != t.size(); ++i) {
if (m.find(t[i]) == m.end())
m[t[i]] = s[i];
else if (m[t[i]] != s[i])
return false;
}
return true;
}
};
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.
class Solution {
public:
bool isIsomorphic(string s, string t) {
map<char, char> m;
for (int i = 0; i != s.size(); ++i) {
if (m.find(s[i]) == m.end())
m[s[i]] = t[i];
else if (m[s[i]] != t[i])
return false;
}
m.clear();
for (int i = 0; i != t.size(); ++i) {
if (m.find(t[i]) == m.end())
m[t[i]] = s[i];
else if (m[t[i]] != s[i])
return false;
}
return true;
}
};
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