LeetCode刷题笔录Longest Valid Parentheses-优快云博客

本文介绍了一种使用栈解决最长有效括号子串问题的方法。通过遍历字符串并利用栈记录左括号的位置，当遇到右括号时计算有效括号子串的长度，并更新最大值。

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.

用一个stack来记录左括号的index。遇到右括号后pop出栈顶得左括号。然后会有以下两种情况：

1.stack变为空。那么length = i - lastInvalidRight;

2.stack不为空。那么length = i - stack.peek();

上面两种情况分别对应的是")()()"以及"(()()"这样子。当然"((()))"也是可以处理的。

public class Solution {
    public int longestValidParentheses(String s) {
        if(s == null || s.length() == 0)
            return 0;
        
        char[] arr = s.toCharArray();
        Stack<Integer> stack = new Stack<Integer>();
        int max = 0;
        int lastInvalidRight = -1;
        for(int i = 0; i < arr.length; i++){
            //push left paren's index onto stack
            if(arr[i] == '('){
                stack.push(i);
            }
            //right paren, pop the first left paren and calculate index difference
            else{
                if(stack.isEmpty()){
                    lastInvalidRight = i;
                    continue;
                }
                else{
                    stack.pop();
                    if(stack.isEmpty())
                        max = Math.max(max, i - lastInvalidRight);
                    else
                        max = Math.max(max, i - stack.peek());
                }
            }
        }
        return max;
    }
}