we set the second last element as the next of the last element and set the next pointer of second last element to NULL. The last element will be marked as the new head. Use the following code to reverse the lin
2.Recursive Method There's code in one reply that spells it out, but you might find it easier to start from the bottom up, by asking and answering tiny questions (this is the approach in The Little Lisper): What is the reverse of null (the empty list)? null. What is the reverse o...
function Node(val) {return{ val, next:null}; } function LinkedList() {return{ head:null, tail:null, add(val) {constnode =newNode(val);if(!this.head) {this.head =node;this.tail =node;returnnode; }this.tail.next =node;this.tail =node;returnnode; },//1 - -2 -- x-- xrevers...
Reverse a linked list 原创转载请注明出处:http://agilestyle.iteye.com/blog/2360694 Recursive Idea Reverse(Head -> Remaining List) => Reverse(Remaining List) -> Head Example: Reverse(1->2->3->4->5) ... 查看原文 Use stack (LIFO) to simulate queue (FIFO)...
publicListNodereverseList(ListNodehead){/* recursive solution */returnreverseListInt(head,null);}privateListNodereverseListInt(ListNodehead,ListNodenewHead){if(head==null)returnnewHead;ListNodenext=head.next;head.next=newHead;returnreverseListInt(next,head);} ...
2. Recursive method: Under the recursive method, all you need to think is the relationship between the head, and the reversed linked list end with head.next. It's pretty clear that what you can do is to put the head at the end of the reversed linked list to make it a reversed linke...
recursive 解法: 总结是传给helper method两个节点,cur和pre(最开始是head和null), 先用n1存当前节点的next,然后把当前节点的next指向pre,然后一直recursively call help method直到过完整个linkedlist. /** * Definition for singly-linked list. * public class ListNode { ...
分析: 以下是Iterative方式的代码 以下是Recursive方式的代码...LinkedList——Reverse Linked List Reverse a singly linked list. Example: Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL Follow up: A linked list can be reversed either iteratively or recursively. ......
//RecursiveclassSolution {public: ListNode* reverseList(ListNode*head) {if(!head || !head->next)returnhead; ListNode*p =head; head= reverseList(p->next); p->next->next =p; p->next =NULL;returnhead; } }; 本文转自博客园Grandyang的博客,原文链接:倒置链表[LeetCode] Reverse Linked List,...
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