if(end !=null)//in case of odd list mid = mid.next; mid = reverseList(mid); while(mid !=null){ if(mid.val != head.val) returnfalse; mid = mid.next; head = head.next; } returntrue; } publicListNode reverseList(ListNode head){ ListNode pre =null, next =null; while(head !=...
Palindrome Linked List Given a singly linked list, determine if it is a palindrome. Follow up: Could you do it in O(n) time and O(1) space? 反转链表 复杂度 时间O(N) 空间 O(1) 思路 两个指针都从头出发,快指针每次两步,慢指针每次一步,这样快指针的下一个或下下个为空时,慢指针就在链...
} if(end!=null)//in case of odd list mid=mid.next; mid=reverseList(mid); while(mid!=null){ if(mid.val!=head.val) returnfalse; mid=mid.next; head=head.next; } returntrue; } publicListNodereverseList(ListNodehead){ ListNodepre=null,next=null; while(head!=null){ next=head.next; ...
由于需要返回两个值,在 Java 中需要使用自定义类进行封装,C/C++ 中则可以使用指针改变在递归调用后进行比较时节点的值。 Java /*** Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * }*/classResult { ListNode lNode;...
Given a list of unique words, find all pairs of distinct indices (i, j) in the given list, so that the concatenation of the two words, i.e. words[i] + words[j] is a palindrome. Given words = ["abcd", "dcba", "lls", "s", "sssll"] ...
234. Palindrome Linked List Given a singly linked list, determine if it is a palindrome. Follow up: Could you do it in O(n) time and O(1) space? /** * Definition for singly-linked list. * public class ListNode { * int val;...
基本是按照九章常规java版的思路写的 DFS 模板, index 指向current 切割到的位置,然后用for 循环查找下一刀切向何处,切完后先测一下是不是回文,不是的话就跳过,是的话继续递归,直到切到字符串结束。 class Solution: """ @param: s: A string @return: A list of lists of string """ def partition(...
Print Palindrome numbers from the given list: In this, we have a list of numbers from that list we have to print only palindrome numbers present in that list, palindrome numbers are numbers, on reversing which number remains the same.
我是python编程的新手,为了解决Spoj挑战,我尝试对回文数字进行编码。我的代码: num_list = list(str(number)) for j in range(len(num_list)): result = ''.join(str(e) for e in elements) 浏览2提问于2017-10-25得票数 0 1回答 文件接受答案而不是打印 、、 我的代码: if i == i[:-1]:...
Could you do it in O(n) time and O(1) space? Hide Tags Linked List Two Pointers 【解题思路】 1、遍历链表,快慢指针,找到链表后半部分。 2、反转链表,能够參考Reverse Linked List 3、然后比較前半部分和后半部分的val值。 Java AC /** ...