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;
    }
};