LeetCode - Longest Valid Parentheses | Darren's Blog

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.

public int longestValidParentheses(String s) {

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

            return 0;

        int longestLength = 0;      // Length of the longest valid parentheses

        int start = 0;  // The start index of the possibly longest valid parentheses

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

        // One-pass scan

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

            if (s.charAt(i) == '(') {   // Opening parenthesis

                stack.push(i);          // Push its index

            } else {        // Closing parenthesis

                if (stack.empty()) {    // No opening parenthesis to match

                    start = i + 1;      // i+1 is the start of next possibly LVP

                } else {

                    stack.pop();    // The index of the opening parenthesis matched by s[i]

                    if (stack.empty())  // s[start...i] is matched

                        longestLength = Math.max(longestLength, i-start+1);

                    else    // s[stack.peek()] is unmatched; s[stack.peek()+1...i] is matched

                        longestLength = Math.max(longestLength, i-stack.peek());

                }

            }

        }

 

        return longestLength;

    }

DP, Code from http://www.darrensunny.me/leetcode-longest-valid-parentheses/

这道题可以用一维动态规划逆向求解。假设输入括号表达式为String s，维护一个长度为s.length的一维数组dp[]，数组元素初始化为0。 dp[i]表示从s[i]到s[s.length - 1]最长的有效匹配括号子串长度。则存在如下关系：

• dp[s.length - 1] = 0;
• 从i - 2 -> 0逆向求dp[]，并记录其最大值。若s[i] == '('，则在s中从i开始到s.length - 1计算s[i]的值。这个计算分为两步，通过dp[i + 1]进行的（注意dp[i + 1]已经在上一步求解）：
  • 在s中寻找从i + 1开始的有效括号匹配子串长度，即dp[i + 1]，跳过这段有效的括号子串，查看下一个字符，其下标为j = i + 1 + dp[i + 1]。若j没有越界，并且s[j] == ‘)’，则s[i ... j]为有效括号匹配，dp[i] =dp[i + 1] + 2。
  • 在求得了s[i ... j]的有效匹配长度之后，若j + 1没有越界，则dp[i]的值还要加上从j + 1开始的最长有效匹配，即dp[j + 1]。

public int longestValidParentheses(String s) {
    // Start typing your Java solution below
    // DO NOT write main() function
    int n = s.length();
    int[] dp = new int[n];
    java.util.Arrays.fill(dp, 0);
    int max = 0;
    for (int i = n - 2; i >= 0; i--) {
      if (s.charAt(i) == '(') {
        int j = i + 1 + dp[i + 1];
        if (j < n && s.charAt(j) == ')') {
          dp[i] = dp[i + 1] + 2;
          int k = 0;
          if (j + 1 < n) {
            k = dp[j + 1];
          }
          dp[i] += k;
        }
        max = Math.max(max, dp[i]);
      }
    }
    return max;
  }
Also check http://yucoding.blogspot.com/2013/01/leetcode-question-46-longest-valid.html
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