LeetCode "Word Pattern II"-优快云博客

本文介绍了一种基于递归的方法来解决字符串模式匹配的问题。通过使用unordered_map存储模式到字符串的映射，该算法能够逐个字符地尝试匹配，并在失败时回溯。此递归解决方案适用于模式和输入字符串长度不固定的情况。

Here is the mind flow: we don't know 1st token right? then we try it one by one - recursively.

typedef unordered_map<string, string> HMap;
class Solution {
    bool go(string pattern, string str, HMap p2s)
    {
        size_t plen = pattern.length();
        size_t slen = str.length();
        if(!plen) return !slen;
        if(plen > slen) return false;

        string c = pattern.substr(0, 1);
        string npat = pattern.substr(1);
        for(int sl = 1; sl <= slen; sl ++)
        {
            string tk = str.substr(0, sl);
            string nstr = str.substr(sl);
            if(p2s.count(c))
            {
                if(p2s[c] == tk)
                {
                    if(go(npat, nstr, p2s))
                        return true;
                }
            }
            else
            {
                HMap np2s = p2s;
                np2s[c] = tk;
                if(go(npat, nstr, np2s))
                    return true;
            }
        }
        return false;
    }
public:
    bool wordPatternMatch(string pattern, string str) {
        size_t plen = pattern.length();
        size_t slen = str.length();
        if(plen == 0) return slen == 0;
        if(plen == 1) return slen > 0;
        if(plen > slen) return false;

        string c = pattern.substr(0, 1);
        string npat = pattern.substr(1);

        for(int sl = 1; sl <= slen; sl ++)
        {
            string head = str.substr(0, sl);
            HMap p2s; p2s[c] = head;

            string nstr = str.substr(sl);
            if(go(npat, nstr, p2s))
                return true;
        }
        return false;
    }
};