290. Word Pattern

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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.

Notes:
You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.

简单版：将str分开，存入数组，再进行pattern循环匹配。

class Solution {public:    bool wordPattern(string pattern, string str) {        vector<string>str_list;        string tmp_s="";        int n=0;        while(true){            n=str.find(" ");            if(n<0){                tmp_s=str.substr(0,str.length());                str_list.push_back(tmp_s);                break;            }            tmp_s=str.substr(0,n);            str.erase(0,n+1);            str_list.push_back(tmp_s);        }        int len=pattern.length();        if(len!=str_list.size())            return false;        for(int i=0;i<len;i++){            for(int j=i+1;j<len;j++){                    if(pattern[i]==pattern[j]){                        if(str_list[i]!=str_list[j]){                            return false;                        }                    }else{                        if(str_list[i]==str_list[j]){                            return false;                        }                    }            }        }        return true;    }};

使用map，建立key-value关系，若key存在则比较值，由于map不能按值查找，所以要对不存在的key查找是否值已经存在，这样效率并不是减少太多

class Solution {public:    bool wordPattern(string pattern, string str) {        vector<string>str_list;        string tmp_s="";        int n=0;        while(true){            n=str.find(" ");            if(n<0){                tmp_s=str.substr(0,str.length());                str_list.push_back(tmp_s);                break;            }            tmp_s=str.substr(0,n);            str.erase(0,n+1);            str_list.push_back(tmp_s);        }        int len=pattern.length();        if(len!=str_list.size())            return false;        map<char,string> maplist;        for(int i=0;i<len;i++){            bool it=maplist.count(pattern[i]);            if(it){                if(str_list[i]!=maplist[pattern[i]])                    return false;            }else{                for(map<char,string>::iterator it=maplist.begin();it!=maplist.end();it++){                    if(it->second==str_list[i])                        return false;                }                maplist.insert(pair<char,string>(pattern[i],str_list[i]));            }        }        return true;    }};

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