LeetCode -- Isomorphic Strings
題目描述:
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.
判斷同構字符串。 就是兩個字符串長度相同,並且每一位被相同的字符替代後所得的新字符串相等,這樣的字符串是同構字符串。
思路:
長度不等直接返回false。
兩個哈希表hash1 和has2。 hash1 : 遍歷第一個字符串的同時,記錄當前字符最後一次出現的位置。
如果hash1中包含了s[i],那麼hash2也一定要包含t[i]並且hash1[s[i]](s[i]最後一次出現的位置)也要等於hash2[t[i]](t[i]最後一次出現的位置)。
如果hash1中沒有包含s[i],那麼hash2也一定不能包含t[i]。
實現代碼:
public class Solution {
public bool IsIsomorphic(string s, string t) {
var hash1 = new Dictionary();
var hash2 = new Dictionary();
for(var i = 0;i < s.Length; i++){
if(!hash1.ContainsKey(s[i])){
hash1.Add(s[i], i);
if(hash2.ContainsKey(t[i])){
return false;
}
hash2.Add(t[i],i);
}
else{
if(!hash2.ContainsKey(t[i]) || hash2[t[i]] != hash1[s[i]]){
return false;
}
hash1[s[i]] = i;
hash2[t[i]] = i;
}
}
return true;
}
}