[leetcode 290]Word Pattern

Given a pattern and a string str, find if str follows the same pattern.

Examples:

Notes:

1. patterncontains only lowercase alphabetical letters, and str contains words separated by a single space. Each word in str contains only lowercase alphabetical letters.
2. Both pattern and str do not have leading or trailing spaces.
3. Each letter in pattern must map to a word with length that is at least 1.

Credits:

Special thanks to  for adding this problem and creating all test cases.

字符与字符串的匹配

AC代码：

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        vector<string> temp=stringTovec(str);
        int len=pattern.size();
        int sum=temp.size();
        map<char,string> m;
        map<string,char> m2;
        int j=0;
        for(int i=0;i<len;++i)
        {
            if(m.find(pattern[i])!=m.end())
            {
                if(j<sum&&m[pattern[i]]==temp[j])
                {
                    ++j;
                    continue;
                }
                else
                    return false;
            }
            else if(j<sum)
            {
                m[pattern[i]]=temp[j];
                ++j;
            }
            else
                return false;
        }
        j=0;
        for(int i=0;i<sum;++i)
        {
            if(m2.find(temp[i])!=m2.end())
            {
               if(m2[temp[i]]==pattern[j])
               {
                   ++j;
                   continue;
               }
               else
                return false;
            }
            else if(j<len)
            {
                m2[temp[i]]=pattern[j];
                ++j;
            }
            else
                return false;
        }
        return true;
    }
private:
    vector<string> stringTovec(string s)
    {
        vector<string> res;
        int len=s.size();
        string temp="";
        for(int i=0;i<len;++i)
        {
            if(s[i]==' ')
            {
                res.push_back(temp);
                temp="";
            }
            else
                temp+=s[i];
        }
        res.push_back(temp);
        return res;
    }
};