LeetCode Longest Valid Parentheses

Description:

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.

Solution:

TLE solution:

we can use dp[i][j] to represent the valid parentheses number.

if( str[i], str[j] is a pair):

dp[i][j] = dp[i+1][j-1] + 1

else:

dp[i][j] = dp[i+1][j-1]

<span style="font-size:18px;">import java.util.*;public class Solution{public int longestValidParentheses(String s) {if (s == null)return 0;int n = s.length();if (n == 0)return 0;char ch1, ch2;int dp[][] = new int[n][n];for (int i = 0; i + 1 < n; i++) {ch1 = s.charAt(i);ch2 = s.charAt(i + 1);if (ch1 == '(' && ch2 == ')')dp[i][i + 1] = 1;}int j;int max = 0;for (int len = 3; len <= n; len++) {for (int i = 0; (j = i + len - 1) < n; i++) {ch1 = s.charAt(i);ch2 = s.charAt(j);if (ch1 == '(' && ch2 == ')') {dp[i][j] = dp[i + 1][j - 1] + 1;} elsedp[i][j] = dp[i + 1][j - 1];max = Math.max(dp[i][j], max);}}return max;}}</span>

Since we get a TLE on this, we may change to another solution, use a stack.

<span style="font-size:18px;">import java.util.*;public class Solution {public int longestValidParentheses(String s) {if (s == null)return 0;int n = s.length();if (n == 0)return 0;int max = 0;Stack<Integer> stack = new Stack<Integer>();int start = 0;for (int i = 0; i < n; i++) {if (s.charAt(i) == '(')stack.push(i);else {if (!stack.isEmpty()) {stack.pop();if (stack.isEmpty())max = Math.max(max, i - start + 1);elsemax = Math.max(max, i - stack.peek() + 1);} else {start = i + 1;}}}return max;}public static void main(String[] args) {Solution s = new Solution();String str = "(()";System.out.println(s.longestValidParentheses(str));}}</span>

Attention to the following two codes, they seem to have same functions, but are completely different. What we need here is the second one. Just feel the difference.

<span style="font-size:18px;">first = stack.pop();if (stack.isEmpty())max = Math.max(max, i - start + 1);elsemax = Math.max(max, i - first + 1);</span>

Second segment of codes

<span style="font-size:18px;">stack.pop();if (stack.isEmpty())max = Math.max(max, i - start + 1);elsemax = Math.max(max, i - stack.peek());</span>