[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.

方法：用stack进行配对，用另一个对应的stack记录配对括号的下标，并把配好对的括号的起始下标记录在名为start的vector里，把对应的结束下标记录在

名为end的vector里，然后遍历两个vector，把能连接起来的括号的数量求出来。

class Solution {
public:
    int longestValidParentheses(string s) {
        int num = 0;
        int big = 0;
        stack<char> st;
        stack<int> si;
        vector<int> start,end;
        int flag = 0;
        for(int i=0;i<s.size();i++){
            if(st.empty()){
                st.push(s[i]);
                si.push(i);
            }             
            else if(s[i]==')' && st.top()=='('){
                    start.push_back(si.top());
                    end.push_back(i);
                    st.pop();
                    si.pop();
            }else{
                  st.push(s[i]);
                  si.push(i);
            } 
        }//end for
        if(start.size()==0)
            return 0;
        while(start.size()>0){

            int min = s.size(),minIndex=0;
            for(int i=0;i<start.size();i++){
                if(start[i]<min){
                    min = start[i];
                    minIndex = i;
                }
            }
            num += (end[minIndex]-start[minIndex]+1);
            int theStart = end[minIndex]+1;

            start.erase(start.begin()+minIndex);
            end.erase(end.begin()+minIndex);
            
            while(find(start.begin(),start.end(),theStart) != start.end()){
                vector<int>::iterator iter1,iter2;
                iter1 = find(start.begin(),start.end(),theStart);
                int d = iter1- start.begin();
                iter2 = end.begin()+d;
                num += ((*iter2)-(*iter1)+1);
                theStart = (*iter2) + 1;
                start.erase(iter1);
                end.erase(iter2);
                
            }//end while
            if(num>big)
                big = num;
            num = 0;
        }
     return big;

    }//end func
};