【一天一道LeetCode】#205. Isomorphic Strings
一天一道LeetCode
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(一)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.
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.
(二)解题
题目大意:给定两个字符串,s中的每个字符能个通过某种映射关系得到t,则返回true,否则返回false
解题思路:如题目给的例子,“egg”和“add”,e->a,g->d,通过这个映射关系可以得到add。
所以,很容易想到用hash表来解决这个问题。不过要注意: 不允许s中的两个不同字符对应t中的同一个字符。
即s = “ab”,t = “aa”,s中a和b不能同时对应t中的a
下面看代码:
class Solution {
public:
bool isIsomorphic(string s, string t) {
int len = s.length();
if(len==0) return true;
//必须要双向映射,避免出现一对多,多对一等情况
char hash[256];
char hash2[256];
memset(hash,' ',256*sizeof(char));
memset(hash2,' ',256*sizeof(char));
for(int i = 0 ; i < len ; i++){
if(hash[s[i]]==' '&&hash2[t[i]]==' '){//如果该组字符没有映射关系
hash[s[i]] = t[i];//建立映射关系
hash2[t[i]] = s[i];
}
else {
if(hash[s[i]]==t[i]&&hash2[t[i]]==s[i]) continue;
else return false;
}
}
return true;
}
};
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