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...
publicListNode reverseList(ListNode head) { ListNode newHead =null; while(head !=null) { ListNode next = head.next; head.next = newHead; newHead = head; head = next; } returnnewHead; } Java: Recursive solution 1 2 3 4 5 6 7 8 9 10 11 publicListNode reverseList(ListNode head) {...
好,分析结束,进入编码阶段。 packagemainimport"fmt"typeListNodestruct{Datainterface{}Next*ListNode}funcReverseListRecursive(head*ListNode)*ListNode{ifhead==nil||head.Next==nil{returnhead}p:=ReverseListRecursive(head.Next)head.Next.Next=headhead.Next=nilreturnp}funcCreateNodeList(node*ListNode,maxint){...
Question Reverse a singly linked list. Solution 1 -- Iterative Remember to set head.next = null or it will report "memory limit exceeds" error. /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } *...
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...
递归拷贝(Recursive Copy) C++中的递归拷贝是一种在拷贝对象时使用递归方法的技术。在C++中,为了拷贝一个对象,通常使用拷贝构造函数或拷贝赋值运算符。然而,对于复杂的对象,拷贝构造函数和拷贝赋值运算符可能无法完全拷贝对象的所有成员。这就需要使用递归拷贝来确保所有成员都能被正确地拷贝。
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 { ...
//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,...
Time: O(n) as we have to traverse all the nodes in the linked list. Space: O(1) as we use a dummy node as the fake head of the reversed list. Recursive solution class Solution { public ListNode reverseList(ListNode head) {
publicListNode reverseList(ListNode head) { ListNode first=null; ListNode tail=head; ListNode tmp;while(tail!=null){ tmp=first; first=tail; tail=tail.next; first.next=tmp; }returnfirst; } 2.Recursive Method There's code in one reply that spells it out, but you might find it easier to...