Longest Valid Parentheses

题目描述

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.

题目解答

解题思想

• 遍历两遍, 第一遍把优先的’(’ 和 ‘)’ 转换成 ’ * ’ ， 然后计算最长连续的*个数
• 动态规划的思想

代码实现

public class Solution {
    public int longestValidParentheses(String s) {
       if(s == null || s.length() == 0)
            return 0;
        StringBuilder temp = new StringBuilder(s);
        ArrayDeque<Integer> stack = new ArrayDeque<>();
        for(int i = 0; i < s.length(); i++){
            if(s.charAt(i) == '(')
                stack.push(i);
            else if(!stack.isEmpty()){
                //注意
                temp.setCharAt(stack.pop(), '*');
                temp.setCharAt(i, '*');
            }
        }

        int longest = 0, part = 0;
        for(int i = 0; i < temp.length(); i++){
            if(temp.charAt(i) == '*'){
                part++;
            }else {
                longest = Math.max(longest, part);
                part = 0;
            }
        }

        //注意
        longest = Math.max(longest, part);
        return longest;
    }
}

DP

public class Solution {
    /** * 1. 状态： DP[i]：以s[i-1]为结尾的longest valid parentheses substring的长度。 2. 通项公式： s[i-1] = '('： DP[i] = 0 s[i-1] = ')'：找i前一个字符的最长括号串DP[i]的前一个字符j = i-2-DP[i-1] DP[i] = DP[i-1] + 2 + DP[j]，如果j >=0，且s[j] = '(' DP[i] = 0，如果j<0，或s[j] = ')' */
    public int longestValidParentheses(String s) {
       if(s == null || s.length() == 0 || s.length() == 1)
            return 0;


        int len = s.length();
        int[] dp = new int[len+1];
        Arrays.fill(dp, 0);

        int maxLen = 0;
        for(int i = 1; i < dp.length; i++){
            int j = i-2-dp[i-1];
            if(s.charAt(i-1) == '(' || j < 0 || s.charAt(j) == ')')
                dp[i] = 0;
            else {
                dp[i] = dp[i-1]+2+dp[j];
                maxLen = Math.max(dp[i], maxLen);
            }
        }

        return maxLen;
    }
}