290. Word Pattern

290. Word Pattern
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.

1. 感觉自己的方法比较笨，不过相对自己以前的编码习惯，在按照空格分割string时，引用istringstream的确是一种比较快捷的方法。

解决这个问题的主要思想：

1. pattern：abba，我们编码为0110；对于pattern中的每个字符，我们将其映射到其第一次出现的位置，这个可以直接利用string的find_first_of得到。
2. str = “dog cat cat fish”，我们编码为0113；对于vector来说，如何实现将每个单词映射到其第一次出现的位置，我引用了map。

class Solution {public:    bool wordPattern(string pattern, string str) {        int* patternCode=new int[pattern.size()];//对pattern进行编码        for(unsigned int i=0;i<pattern.size();i++)            patternCode[i]=pattern.find_first_of(pattern[i]);        vector<string> vec;        map<string,int> strMap;        istringstream ss(str);        string word;        while(ss>>word)            vec.push_back(word);//分割str为单词，存储在vector中        int* strCode=new int[vec.size()];        for(unsigned int i=0;i<vec.size();i++)//为str进行编码        {            if(!strMap.count(vec[i]))            {                strMap[vec[i]]=i;                strCode[i]=i;            }            else                strCode[i]=strMap[vec[i]];        }        if(pattern.size()!=vec.size())//两者编码长度不同，返回错误            return false;        for(unsigned int i=0;i<vec.size();i++)            if(patternCode[i]!=strCode[i])                return false;        return true;    }};