英文:
Palindrome Linkedlist using recursion
问题
索引从0开始。
输入格式：链表元素（用空格分隔，以-1结尾）
某个隐藏的测试用例出现了运行时错误。 main() 函数正在输入并传递 head。
public class Solution {
static String s1 = "", s2 = "";
public static boolean isPalindrome_2(LinkedListNode<Integer> head) {
if (head != null) {
s1 = s1 + head.data;
isPalindrome_2(head.next);
s2 = s2 + head.data;
}
if (s1.equals(s2))
return true;
return false;
}
}
算法的工作原理：s1 将存储包含所有数据的字符串。 s2 将以相反的方式存储数据，因为它在递归函数之后。然后可以比较这两个字符串。
英文:
Indexing starts from 0.
Input format : Linked list elements (separated by space and terminated by -1)
There is some runtime error on one hidden test case. main() is taking input and passing head.
public class Solution {
static String s1="",s2="";
public static boolean isPalindrome_2(LinkedListNode<Integer> head) {
if (head != null) {
s1 = s1 + head.data;
isPalindrome_2(head.next);
s2 = s2 + head.data;
}
if (s1.equals(s2))
return true;
return false;
}
}
How the algorithm is supposed to work: s1 will store the string containing all data. s2 will store data in reverse manner since it is after recursive function. Then the strings can be compared.
答案1
得分: 0
由于您没有提供可复现的示例，我无法运行您的代码并验证任何假设。因此，以下内容仅基于检查得出。
- 如果列表中的数字是 1 和 11，它不是一个回文。然而，您的字符串
s1和s2都应该变为111，它们是相等的，所以您的方法应该返回true。 - 您没有测试末尾的 -1。难道不应该测试吗？
另外，我没有看到任何索引，也没有看到元素之间有任何空格。
我也没有理解运行时错误是如何发生的，抱歉。再次运行您的代码可能会使我有更大的机会理解。
英文:
Since you have not provided a reproducible example, I have not been able to run your code and verify any hypotheses. So this is from inspection alone.
- If the numbers in the list ar 1 and 11, it is not a palindrome. Yet your string
s1ands2should both become111, that is equal, so your method should returntrue. - You are not testing for the -1 at the end. Shouldn’t you?
BTW I don’t see any indexing and I don’t see any space between elements.
I also have not understood how a runtime error can occur, sorry. Again, running your code might have given me a greater chance.
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