leetcode problem 32 -- Longest Valid Parentheses

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.

 

 

思路1：

　　从左到右依次扫描，用一个数组v记录括号是否已经匹配。然后再扫描一次v，用一个变量count记录连续合法括号数目。如果v[i] = 0则说明该括号未匹配，合法括号序列终端，count = 0. 代码如下：

Runtime: 22 ms

class Solution {

public:

    int longestValidParentheses(string s) {

        vector<int> v(s.length(), 0);

        stack<int> stac;

        for (int i = 0; i < s.length(); ++i) {

            if (s[i] == ')') {

                if (stac.empty())

                    continue;

                else {

                    v[i] = 2;

                    v[stac.top()] = 2;

                    stac.pop();

                }

            }

            else 

                stac.push(i);

        }



        int res = 0, count = 0;

        for (int i = 0; i < s.length(); ++i) {

            if (v[i] == 0) {

                if (res < count)

                    res = count;

                count = 0;

            }

            else if (s[i] == '(')

                count += v[i];

        }

        return res < count ? count : res;

    }



};

 

思路二：

　　先从左到右扫描，分别记录'('和')'的数目， 如果一旦')'的数目大于'(' 那么可以确定合法连续括号序列中断。记录countMax。  

　　然后从右往左扫描，同样分别记录'('和')'的数目， 如果一旦'('的数目大于')' , 那么可以确定合法连续括号序列中断。记录countMax。  

代码如下：

Runtime: 10 ms

class Solution {

public:

    int longestValidParentheses(string s) {

        int ll = 0, lr = 0, li = 0;    

        int rl = 0, rr = 0, ri = s.length()-1;

        int res = 0;

        for (; li < s.length() && ri >= 0; ++li, --ri) {

            switch (s[li]) {

                case '(':

                    ++ll;

                    break;

                case ')':

                    ++lr;

            }

            switch (s[ri]) {

                case '(':

                    ++rl;

                    break;

                case ')':

                    ++rr;

            }

            if (ll == lr && (ll * 2) > res) 

                res = 2 * ll;

            else if (ll < lr)

                ll = lr = 0;

            

            if (rl == rr && (rl * 2) > res) 

                res = 2 * rl;

            else if (rl > rr)

                rl = rr = 0;

            

        }

        

        return res;

    }

private:



};