[LeetCode]Longest Valid Parentheses

class Solution {//once meet Parentheses, we should think of stack//then it is some thing trivialpublic:int longestValidParentheses(string s) {// Start typing your C/C++ solution below// DO NOT write int main() functionstack<pair<char, int>> charStack;vector<bool> valid(s.size(), false);for (int i = 0; i < s.size(); ++i){if ( !charStack.empty() ){char topChar = charStack.top().first;int topPos = charStack.top().second;if( topChar == '(' && s[i] == ')'){charStack.pop();for (int j = topPos; j <= i; ++j)valid[j] = true;}else charStack.push(make_pair(s[i], i));}else charStack.push(make_pair(s[i], i));}//then count the longest lengthint nowLen = 0;int maxLen = 0;for (int i = 0; i < s.size(); ++i){if(valid[i])nowLen++;else {maxLen = max(maxLen, nowLen);nowLen = 0;}}maxLen = max(maxLen, nowLen);//note here, then end of the stringreturn maxLen;}};

second time

class Solution {//DP based solution need O(n^3)//stack based solution need O(n^2)public:    struct Node    {        char c;        int pos;        Node(char _c, int _pos):c(_c),pos(_pos){};    };    int longestValidParentheses(string s) {        // Start typing your C/C++ solution below        // DO NOT write int main() function       stack<Node> charStack;       vector<int> pLen(s.size(), 0);       for(int i = 0; i < s.size(); ++i)       {           if(s[i] == '(') charStack.push(Node(s[i], i));           else           {               if(!charStack.empty() && charStack.top().c == '(')               {                   int pos = charStack.top().pos;                   pLen[pos] = i-pos+1;                   charStack.pop();               }               else charStack.push(Node(s[i], i));           }       }       //get maxlen       int maxLen = 0;       for(int i = 0; i < s.size(); ++i)       {           int curLen = 0;           int nextPos = i;           while(nextPos < s.size() && pLen[nextPos] != 0)           {               curLen += pLen[nextPos];               nextPos = pLen[nextPos]+nextPos;           }           maxLen = max(maxLen, curLen);       }              return maxLen;    }};