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_32.java
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_32.java
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package com.fishercoder.solutions;
import java.util.Stack;
public class _32 {
public static class Solution1 {
public int longestValidParentheses(String s) {
int result = 0;
Stack<Integer> stack = new Stack();
stack.push(-1);
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '(') {
stack.push(i);
} else {
stack.pop();
if (stack.isEmpty()) {
stack.push(i);
} else {
result = Math.max(result, i - stack.peek());
}
}
}
return result;
}
}
public static class Solution2 {
/**
* my lengthy but original solution on 4/5/2021, the idea is to convert the valid parenthesis pairs into numbers and push them onto a stack.
*/
public int longestValidParentheses(String s) {
Stack<String> stack = new Stack<>();
int longest = 0;
for (char c : s.toCharArray()) {
if (c == '(') {
stack.push(c + "");
} else {
if (stack.isEmpty()) {
continue;
} else {
if (stack.peek().equals("(")) {
stack.pop();
stack.push("2");
} else {
int sum = 0;
while (!stack.isEmpty() && !stack.peek().equals("(")) {
sum += Integer.parseInt(stack.pop());
}
if (!stack.isEmpty()) {
stack.pop();//pop off the open paren
sum += 2;
stack.push("" + sum);
}
longest = Math.max(longest, sum);
}
}
}
}
int result = 0;
while (!stack.isEmpty()) {
if (stack.peek().equals("(")) {
stack.pop();
result = 0;
} else {
result += Integer.parseInt(stack.pop());
longest = Math.max(result, longest);
}
}
return longest;
}
}
}