[leetcode] 205.Isomorphic Strings

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.

题意：

判断一个字符串与两一个字符串是否是重构的，就是一个字符串可以转换到另一个字符串，通过改变每个字符。但是同一个字符不能转为多个字符。

思路：

使用hash来完成，考虑到字符个数比较少，可以直接使用数组来实现hash的功能。不过需要从s到t的转变与t到s的转变都查看，比如ab->aa，从s到t查看是没问题，但从t到s的转变我们发现a指向了a与b。所以两者不是同构。同理aa->ab也不是同构。

以上所述。

代码如下：

class Solution {
public:
bool isIsomorphic(string s, string t) {
if(s.length() != t.length())return false;
int s2t[128];
int t2s[128];
memset(s2t, -1, sizeof(s2t));
memset(t2s, -1, sizeof(t2s));
int len = s.length();
for(int i = 0; i < len; i++)
{
if(s2t[s[i]] == -1)
{
s2t[s[i]] = t[i];
}
if(t2s[t[i]] == -1)
{
t2s[t[i]] = s[i];
}
if(s2t[s[i]] != t[i] || t2s[t[i]] != s[i])
return false;
}
return true;
}
};