Isomorphic Strings——同构结构判断

问题描述：求两个字符串是否同构
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.

思路：
这里用到的工具是unordered_map。
设置两个map1，map2，分别对两个字符串作为key进行对应的赋值，如果出现一个map中的值不等于对应的map的键值，返回false，当所有元素对应，两个字符串同构。

程序如下：

#include<iostream>#include <vector>#include<string>#include<unordered_map>using namespace std;class Solution {public:    bool isIsomorphic(string s, string t) {        if (s.size() != t.size())            return false;        unordered_map<char,char> map1;        unordered_map<char, char> map2;        for (int i = 0;i < s.size();i++)        {            if ((map1.find(s[i]) == map1.end())&&(map2.find(t[i])==map2.end()))            {                map1[s[i]] = t[i];                map2[t[i]] = s[i];            }            else if (map1[s[i]] != t[i]||map2[t[i]]!=s[i])                return false;        }        return true;    }};void main(){    Solution so;    string s = "egg";    string t = "absb";    cout << so.isIsomorphic(s, t) << endl;}

再深入一点，word pattern问题，如下：
Given a pattern and a string str, find if str follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.

Examples:
pattern = “abba”, str = “dog cat cat dog” should return true.
pattern = “abba”, str = “dog cat cat fish” should return false.
pattern = “aaaa”, str = “dog cat cat dog” should return false.
pattern = “abba”, str = “dog dog dog dog” should return false.

同样方法类似：
这里还要用到一个工具：stringstream ss(str);头文件#include<sstream>
把字符串按空格分开，ss>>a;赋值给a；

程序如下：

class Solution {public:    bool wordPattern(string pattern, string str) {        unordered_map<char,string> map1;        unordered_map<string, char> map2;        string a;        stringstream ss(str);        for (auto c : pattern)        {            if (!(ss >> a))            {                return false;            }            if (map1.find(c) == map1.end() && map2.find(a) == map2.end())            {                map1[c] = a;                map2[a] = c;            }            if (map1[c] != a || map2[a] != c)            {                return false;            }        }            if (!ss.eof())             {                return false;            }        return true;    }};