LeetCode:Word Pattern

问题描述：

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.

Credits:
Special thanks to @minglotus6 for adding this problem and creating all test cases.

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思路：

1、先将目标串分割；

2、判断目标串和分割后的串的长度是否相等；

3、使用map自带函数判断是否匹配；

代码：

class Solution {
public:
    bool wordPattern(string pattern, string str) {
    vector<string> src;
    string word = "";
    for(int i = 0;i < str.size();++i)
    {
        if(str[i] == ' ')
        {
            src.push_back(word);
            word = "";
        }
        else
        {
            word += str[i];
        }
        if(i == str.size() - 1)
        {
            src.push_back(word);
        }
    }
    int n = pattern.size();
    if(n != src.size())
        return false;
        
    map<char,string> mp1;
    map<string,char> mp2;
    for(int i = 0;i < n; ++i)
    {
        map<char,string>::iterator mp1iter;
        map<string,char>::iterator mp2iter;
        mp1iter = mp1.find(pattern[i]);
        mp2iter = mp2.find(src[i]);
        if(mp1iter == mp1.end() && mp2iter == mp2.end())
        {
            mp1[pattern[i]] = src[i];
            mp2[src[i]] = pattern[i];
        }
        else if(mp1iter != mp1.end() && mp2iter != mp2.end())
             {
                 if(mp1[pattern[i]] != src[i] || mp2[src[i]] != pattern[i])
                    return false;
             }
             else
                return false;
    }
    return true;   
    }
};