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

;
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;
} else
dp[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);
else
max = 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);
else
max = 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);
else
max = Math.max(max, i - stack.peek());</span>