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.

Example 1:

Input: s = [code]"egg",

t =

"add"

Output: true [/code]

Example 2:

Input: s = [code]"foo",

t =

"bar"

Output: false[/code]

Example 3:

Input: s = [code]"paper",

t =

"title"

Output: true[/code]

Note:
You may assume both s and t have the same length.

给定两个字符串s和t，判断它们是否是同构的。如果字符串s可以通过字符替换的方式得到字符串t，则称s和t是同构的。字符的每一次出现都必须被其对应字符所替换，同时还需要保证原始顺序不发生改变。两个字符不能映射到同一个字符，但是字符可以映射到其本身。假设两个字符串长度相等。 

解法1: 将字符串全部输出成数字序列， Given "paper", return "01023". Given "isomorphic", return "0123245607". 最后在比较两个数字是否相等。超时不能Accept。

解法2: 哈希表，用哈希表记录每一个字符出现的位置，如果对应的字符出现的位置不一样，就返回False，到最后都一样就返回True。也可以只记录最后一个字符出现的位置，因为之前出现的位置已经比较过了。

Java:

public class Solution {
public boolean isIsomorphic(String s1, String s2) {
int[] m = new int[512];
for (int i = 0; i < s1.length(); i++) {
if (m[s1.charAt(i)] != m[s2.charAt(i)+256]) return false;
m[s1.charAt(i)] = m[s2.charAt(i)+256] = i+1;
}
return true;
}
}　　

Java:

class Solution {
public boolean isIsomorphic(String s, String t) {
int[] a = new int[256];
int[] b = new int[256];
for (int i = 0; i < 256; i++) {
a[i] = b[i] = -1;
}
int len = s.length();
for (int i = 0; i < len; i++) {
if (a[s.charAt(i)] != b[t.charAt(i)]) {
return false;
}
a[s.charAt(i)] = b[t.charAt(i)] = i;
}

return true;
}
}　　

Python: wo

class Solution(object):
def isIsomorphic(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
ms, mt = {}, {}
for i in xrange(len(s)):
if ms.get(s[i]) != mt.get(t[i]):
return False
ms[s[i]] = i
mt[t[i]] = i

return True   　　

Python:

def isIsomorphic1(self, s, t):
d1, d2 = {}, {}
for i, val in enumerate(s):
d1[val] = d1.get(val, []) + [i]
for i, val in enumerate(t):
d2[val] = d2.get(val, []) + [i]
return sorted(d1.values()) == sorted(d2.values())

def isIsomorphic2(self, s, t):
d1, d2 = [[] for _ in xrange(256)], [[] for _ in xrange(256)]
for i, val in enumerate(s):
d1[ord(val)].append(i)
for i, val in enumerate(t):
d2[ord(val)].append(i)
return sorted(d1) == sorted(d2)

def isIsomorphic3(self, s, t):
return len(set(zip(s, t))) == len(set(s)) == len(set(t))

def isIsomorphic4(self, s, t):
return [s.find(i) for i in s] == [t.find(j) for j in t]

def isIsomorphic5(self, s, t):
return map(s.find, s) == map(t.find, t)

def isIsomorphic(self, s, t):
d1, d2 = [0 for _ in xrange(256)], [0 for _ in xrange(256)]
for i in xrange(len(s)):
if d1[ord(s[i])] != d2[ord(t[i])]:
return False
d1[ord(s[i])] = i+1
d2[ord(t[i])] = i+1
return True　　

C++:

class Solution {
public:
bool isIsomorphic(string s, string t) {
int m1[256] = {0}, m2[256] = {0}, n = s.size();
for (int i = 0; i < n; ++i) {
if (m1[s[i]] != m2[t[i]]) return false;
m1[s[i]] = i + 1;
m2[t[i]] = i + 1;
}
return true;
}
};

　　

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