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.

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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;    }};