[LeetCode]Longest Valid Parentheses

最新推荐文章于 2022-09-01 00:56:47 发布

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本文介绍了两种方法解决C++中寻找字符串中最长有效括号子串的问题：一种是使用栈实现，另一种是基于动态规划。详细解释了每种方法的思路和代码实现。

class Solution {
//once meet Parentheses, we should think of stack
//then it is some thing trivial
public:
	int longestValidParentheses(string s) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		stack<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 length
		int 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 string
		return 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;
    }
};