题目描述

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.

Input: pattern = “abba”, str = “dog cat cat dog”
Output: true

Input:pattern = “abba”, str = “dog cat cat fish”
Output: false

Input: pattern = “aaaa”, str = “dog cat cat dog”
Output: false

Input: pattern = “abba”, str = “dog dog dog dog”
Output: false

解题思路

将两个字符串全部映射到数字数组中，然后比对数字数组
代码如下：

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        int len = pattern.size();
        vector<int> v,vc;
        map<string,int> m;
        for(int i =0;i<len;i++)
        {
            if(pattern.find(pattern[i])!=string::npos && pattern.find(pattern[i])<i)
                v.push_back(pattern.find(pattern[i]));
            else 
                v.push_back(i);
        }
        string sub;
        istringstream is(str);
        int cnt=0;
        while(is>>sub)
        {
            map<string,int>::iterator iter;
            iter = m.find(sub);
            if(iter!=m.end())
                vc.push_back(iter->second);
            else 
                vc.push_back(cnt);
            m.insert(make_pair(sub,cnt++));
        }
        if(cnt!=len) return false;
        for(int i=0;i<v.size();i++)
        {
            if(v[i]!=vc[i])
                return false;
        }
        return true;
    }
};

但是可以继续优化，直接字符串与字符进行映射
代码如下:

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        int len = pattern.size();
        map<string,char> m;
        string sub;
        istringstream is(str);
        int cnt=0;
        while(is>>sub)
        {
            multimap<string,char>::iterator iter;
            iter = m.find(sub);
            if(iter!=m.end())
            {
                if(iter->second != pattern[cnt])   //如果该字符串之前匹配了但是其映射和pattern中的不一样
                    return false;
            }
            else
            {
                //如果sub不在map中，但sub在pattern中的映射与另一个sub一样
                if(pattern.find(pattern[cnt]) !=string::npos && pattern.find(pattern[cnt])<cnt)
                    return false;
            }
            m.insert(make_pair(sub,pattern[cnt++]));
        }
        if(cnt!=len)
             return false;
        return true;
    }
};