leetcode 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:

1. pattern = "abba", str = "dog cat cat dog" should return true.
2. pattern = "abba", str = "dog cat cat fish" should return false.
3. pattern = "aaaa", str = "dog cat cat dog" should return false.
4. 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. 

这个题和判断连个字符串是否同构是类似的，会那一道这一道就不成问题了。

只是多了一个字符串分割，我自己写了一个关于字符串分割的函数。除此之外需要构建两个map，一个是map<string,char>一个是map<char,string>

然后就是没有什么难度了，有一个问题就是我在vs2010里mp[pattern[i]]!=v1[i]这句提示不能用“!=”(原来我是没有#include<string>太粗心啦)不过在leetcode的编译器里通过了。

我需要看一下关于字符串的比较。

class Solution {
public:
    vector<string> splitString(string s){
        vector<string> result;
        string temp;
        int length=s.length();
        for(int i=0;i<length;i++){
            if(s[i]!=' '&&i!=length-1) temp+=s[i];
            else {
                if(i==length-1) temp+=s[i];
                result.push_back(temp);
                temp="";
            }
        }
        return result;
    }
    
    
    bool wordPattern(string pattern, string str) {
        vector<string> v1;
        v1=splitString(str);
        if(pattern.length()!=v1.size()) return false;
        map<char,string> mp;
        for(int i=0;i<pattern.length();i++){
            if(mp.find(pattern[i])==mp.end()) mp[pattern[i]]=v1[i];
            else if(mp[pattern[i]]!=v1[i]) return false;
        }
        map<string,char> mp2;
        for(int i=0;i<pattern.length();i++){
        if(mp2.find(v1[i])==mp2.end()) mp2[v1[i]]=pattern[i];
            else if(mp2[v1[i]]!=pattern[i]) return false;
        }
        return true;
    }
};

这是别人这道题分割字符串的方法，学习了：

        while (sin >> tmp) dic.push_back(tmp);
        if (dic.size() != pattern.size()) return false;