leetcode Longest Valid Parentheses

Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring.

For "(()", the longest valid parentheses substring is "()", which has length = 2.

Another example is ")()())", where the longest valid parentheses substring is "()()", which has length = 4.

You have to consider every possibility, like "(()()", "()((()()", etc

two brilliant O(n) solution

one uses stack, one in DP

class Solution {
public:
    int longestValidParentheses(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function

    int nMax = 0;
    const char* p = s.c_str();
    const char *str=p;
    stack<const char*> stk;
    while (*p )
    {
        if (*p == '(')
            stk.push(p);
        else if (*p == ')')
        {
            if (!stk.empty() && *stk.top() == '(')
            {
                stk.pop();//first pop
                nMax = max(p - (stk.empty() ? str-1 : stk.top()), nMax);//then count by the last remaining
            }
            else stk.push(p);
        }

        p++;
    }

    return nMax;
    }
};

the insight here is to keep the last remaining Parentheses for counting

This DP is definitely not easy to think of. It needs very deep insight.

public class Solution {
    public int longestValidParentheses(String s) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(s==null||s.length()==0) return 0;
        int l=s.length();
        int dp[]=new int[l];
        int max=0,j;
        
        for(int i=1;i<l;i++){
              if(s.charAt(i)==')'){ j=i-1-dp[i-1];
              if(j>=0&&s.charAt(j)=='(') dp[i]=i-j+1+((j>0)?dp[j-1]:0);}

              if(dp[i]>max) max=dp[i];      
              }
              return max;
    }
            }