Title description
http://oj.ecustacm.cn/problem.php?id=1321
A simple regular expression: a regular expression consisting only of x () |.
Xiao Ming wants to find the length of the longest string that this regular expression can accept.
For example ((xx|xxx)x|(x|xx))xx The longest string that can be accepted is: xxxxxx, the length is 6
Explanation:
Regular expressions, also known as regular expressions, are usually used to retrieve and replace text that meets a certain pattern (rule).
For example, for the regular expression composed of x () | in the title, the parenthesis "()" has the highest priority, or the operation "|" takes the second place. Inside the brackets is a whole, or keep the longest one on both sides .
((xx|xxx)x|(x|xx))How is xx executed and why is it 6?
Execute the parentheses first, and then execute the or. The steps are:
(1) Look at the first parenthesis and find that there are nested parentheses inside. Find the innermost parenthesis. Inside the parentheses is an OR operation. ( (xx|xxx) x|(x|xx))xx, get: ( xxx x|(x|xx))xx
(2) Continue to execute the innermost bracket. (xxxx| (x|xx) )xx, get: (xxxx| xx )xx
(3) Continue to execute the last bracket. (xxxx|xx) xx, get: xxxx xx, end, get a string of length 6.
Regular expression tutorial: https://www.runoob.com/regexp/regexp-tutorial.html , it comes with its own testing tool.
answer:
https://blog.csdn.net/weixin_43914593/article/details/112363933
Ni Wendi's words: "This problem is done with a recursive idea, and the code is quite concise. When you encounter a left bracket'(', go to recursive solution, when you encounter a right bracket')', the current recursion ends and the result is returned; |'then record and reset the current value, and then calculate and update to the right; if it is x, just count directly."
while(pos < len){
if (s[pos] =='('){ //Left parenthesis, continue recursion. It is equivalent to stacking
pos++;
tmp += dfs();
} else if (s[pos] ==')'){ //Close parenthesis, return recursively. Equal to pop
pos++;
break;
} else if (s[pos] =='|'){ //check or
pos++;
ans = max (ans, tmp);
tmp = 0;
} else {//Check X and count the number of X
pos++;
tmp++;
}
}
ans = max (ans, tmp);
return ans;
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<set>
#include<queue>
#include<stack>
#include<map>
using namespace std;
typedef long long ll;
const int maxn = 110000;
string s;
int pos=0;//字符串当前指针
int dfs()
{
int maxx=0;
int ans=0;
int len=s.size();
while(pos<len)
{
if(s[pos]=='(')
{
pos++;
maxx+=dfs();//模仿入栈，继续递归
}
else if(s[pos]==')')
{
pos++;
break;//模仿出栈，结束循环
}
else if(s[pos]=='|')
{
pos++;
ans=max(ans,maxx);//取得最大X，为了统计｜之后的X的数量，maxx清零
maxx=0;
}
else if(s[pos]=='x')
{
pos++;
maxx++;//统计X个数
}
}
ans=max(ans,maxx);
return ans;//返回本层递归最大X个数
}
int main()
{
cin>>s;
cout<<dfs();
return 0;
}