Isomorphic Strings(leetcode 205)

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.

思路：

1.提取模式，对比模式

2, 就是记录遍历s的每一个字母，并且记录s[i]到t[i]的映射，当发现与已有的映射不同时，说明无法同构，直接return false。但是这样只能保证从s到t的映射，不能保证从t到s的映射，所以交换s与t的位置再重来一遍上述的遍历就OK了。

3. 建一个map保存映射关系, 同时用一个set保持 被映射的char, 保证同一个char 不会被映射两次.

4. ，依次用‘0’， ‘1‘…替换字符串出现的字符，如‘abc’替换为’012‘， ’abbc‘替换成’0112‘。所以需要设置一张转换表，记录转换后每个字符对应的替代字符。

5.

class Solution {
public:
bool isIsomorphic(string s, string t) {
return getMode(s) == getMode(t);
}
vector<int> getMode(string s) {
vector<int> r;
map<char, int> t;   // int = i + 1;
for (int i = 0; i < s.size(); ++i) {
if (t[s[i]] == 0) {
t[s[i]] = i + 1;
r.push_back(0);
} else {
r.push_back(t[s[i]]);
}
}
return r;
}
};

-----------
class Solution {
public:
bool isIsomorphic(string s, string t) {
if (s.length() != t.length()) return false;
map<char, char> mp;
for (int i = 0; i < s.length(); ++i) {
if (mp.find(s[i]) == mp.end()) mp[s[i]] = t[i];
else if (mp[s[i]] != t[i]) return false;
}
mp.clear();
for (int i = 0; i < s.length(); ++i) {
if (mp.find(t[i]) == mp.end()) mp[t[i]] = s[i];
else if (mp[t[i]] != s[i]) return false;
}
return true;
}
};
---------
public class Solution {
//test case: "egg", "add"

public boolean isIsomorphic(String s, String t) {
//init check
if(s==null || t==null) return false;
if(s.length() != t.length()) return false;

Map<Character, Character> map = new HashMap<Character, Character>();
Set<Character> set = new HashSet<Character>();

for(int i=0; i<s.length(); i++) {
char c1 = s.charAt(i);
char c2 = t.charAt(i);

if(map.containsKey(c1)) {
if(map.get(c1) != c2) return false;
} else {
if(set.contains(c2)) return false;
else {
map.put(c1, c2);
set.add(c2);
}
}
}
return true;
}
}
-----------
4
class Solution {
public:
string transferStr(string s){
char table[128] = {0};
char tmp = '0';
for (int i=0; i<s.length(); i++) {
char c = s.at(i);
if (table[c] == 0) {
table[c] = tmp++;
}
s[i] = table[c];
}
return s;
}
bool isIsomorphic(string s, string t) {

if (s.length() != t.length()) {
return false;
}
if (transferStr(s) == transferStr(t)) {
return true;
}
return false;
}
};