Longest Valid Parentheses

Longest Valid Parentheses

问题：

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：

　　栈的经典应用

我的代码：

public class Solution {

    public int longestValidParentheses(String s) {

        if(s==null || s.length()==0)    return 0;

        Stack<Integer> stack = new Stack<Integer>();

        int max = 0;

        

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

        {

            if(stack.isEmpty() || s.charAt(stack.peek())==')' || s.charAt(i)=='(') stack.push(i);

            else

            {

                stack.pop();

                int index = stack.isEmpty()? -1 : stack.peek();

                max = Math.max(max,i-index);

            }

        }

        return max;

    }

}

View Code

他人代码：

public int longestValidParentheses(String s) {

    char[] S = s.toCharArray();

    int[] V = new int[S.length];

    int open = 0;

    int max = 0;

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

        if (S[i] == '(') open++;

        if (S[i] == ')' && open > 0) {

            V[i] = 2 + V[i-1] + (i-2-V[i-1] > 0 ? V[i-2-V[i-1]] : 0);

            open--;

        }

        if (V[i] > max) max = V[i];

    }

    return max;

}

View Code

学习之处：

• 不知道这道题为什么在leetcode上hard的难度，思路很容易想，代码很容易些，也一遍就AC了
• 想的越多，最后写出来的代码越简洁
• 看别人的代码才发现，这道题的另外一个解法使用动态规划来解， V[i-1] is its previous consecutive parentheses, i-2-V[i-1] is the position right outside this big "( ... )" part. So check if previous big "(...)" is a valid one. 
• 改掉不好的习惯 day by day