算法复习：递归

leedcode 10. 正则表达式匹配

递归解法，时间复杂度比较高，后面要尝试改成动规

bool end_or_not(string p)
{
    for(int i=0;i<p.size();i+=2)
    {
        if(((p[i]>='a'&&p[i]<='z')||p[i]=='.')&&p[i+1]=='*')
            continue;
        return false;
    }
    return true;
}
bool check(string s, string p)
{
    int donser[27];
    memset(donser,0,sizeof(donser));
    for(int i=0;i<p.size();i++)
    {
        if(p[i]=='.'||p[i]=='*')
            continue;
        if(donser[p[i]-'a']==0)
        {
            donser[p[i]-'a']=1;
            int lable=0;
            if(p[i+1]=='*')
                continue;
            for(int j=0;j<s.size();j++)
            {
                if(s[j]==p[i])
                {
                    lable=1;
                    break;
                }
            }
            if(lable==0)
                return false;
        }
    }
    return true;
}
class Solution {
public:
   bool isMatch(string s, string p)
    {
        int ptr_s=0,ptr_p=0;
        bool x=0;
        char now=p[ptr_p],next=p[ptr_p+1];
        if(check(s,p)==false)
            return false;
        if(s[ptr_s]=='\0')
        {
            return end_or_not(p.substr(ptr_p,p.size()-ptr_p));
        }
        if(p[ptr_p]=='\0')
            return false;
        if((now>='a'&&now<='z'&&next!='*')||(now=='.'&&next!='*'))
        {
            if(s[ptr_s]==p[ptr_p]&&s[ptr_s+1]=='\0'&&p[ptr_p+1]=='\0')
                return true;
            if(s[ptr_s]==p[ptr_p]||(now=='.'&&next!='*'))
            {
                ptr_s++;
                ptr_p++;
                x=isMatch(s.substr(ptr_s,s.size()-ptr_s),p.substr(ptr_p,p.size()-ptr_p));
                ptr_s--;
                ptr_p--;
                if(x==true)
                    return true;
            }
            else
            {
                return false;
            }
        }
        if(now=='.'&&next=='*')
        {
            ptr_p+=2;
            x=isMatch(s.substr(ptr_s,s.size()-ptr_s),p.substr(ptr_p,p.size()-ptr_p));
            ptr_p-=2;
            if(x==true)
                return true;
            ptr_s++;
            x=isMatch(s.substr(ptr_s,s.size()-ptr_s),p.substr(ptr_p,p.size()-ptr_p));
            ptr_s--;
            if(x==true)
                return true;
        }
        if(now>='a'&&now<='z'&&next=='*')
        {
            if(s[ptr_s]!=p[ptr_p])
            {
                ptr_p+=2;
                x=isMatch(s.substr(ptr_s,s.size()-ptr_s),p.substr(ptr_p,p.size()-ptr_p));
                ptr_p-=2;
                if(x==true)
                    return true;
            }
            else
            {
                ptr_p+=2;
                ptr_s++;
                x=isMatch(s.substr(ptr_s,s.size()-ptr_s),p.substr(ptr_p,p.size()-ptr_p));
                ptr_p-=2;
                ptr_s--;
                if(x==true)
                    return true;
                ptr_s++;
                x=isMatch(s.substr(ptr_s,s.size()-ptr_s),p.substr(ptr_p,p.size()-ptr_p));
                ptr_s--;
                if(x==true)
                    return true;
                ptr_p+=2;
                x=isMatch(s.substr(ptr_s,s.size()-ptr_s),p.substr(ptr_p,p.size()-ptr_p));
                ptr_p-=2;
                if(x==true)
                    return true;
            }
        }
        return false;
    }
};

 

来源：https://www.cnblogs.com/dzzy/p/12274354.html