开始处理每一个节点ListNode second = first.next;//先处理第一个节点first,所以需要一个指针来存储first的后继first.next = reverseHead;//将first放到新链表头节点的头部reverseHead = first;//移动新链表的头指针,让它始终指向新链表头部first = second;//继续处理原链表的节点,即之前指针
*/classSolution{publicListNodereverseList(ListNode head){//1.基本问题的解if(head ==null|| head.next ==null){returnhead; }//2.将大问题分解成小问题ListNodereve=reverseList(head.next);//3.将小问题的解变成大问题的解head.next.next = head; head.next =null;returnreve; } } 对代码进行解释:...
publicclassSolution{publicListNodeReverseList(ListNode head){if(head==null)returnnull;//head为当前节点,如果当前节点为空的话,那就什么也不做,直接返回null;ListNode pre=null;ListNode nextnode=null;while(head!=null){nextnode=head.next;head.next=pre;pre=head;head=nextnode;}returnpre;}} Reverse Lin...
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1->2->3->4->5->NULL, m = 2 and n = 4, return1->4->3->2->5->NULL. Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list. 迭代法 复杂度 ...
Reverse a singly linked list. click to show more hints. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 既然问了能否iteratively or recursively, 那就both把. iterative 解法: 总结就是得到下一个节点,更改当前节点指向,将指针往下移动,直到过完整个linked...
Reverse a singly linked list. 反转链表,我这里是采用头插法来实现反转链表。 代码如下: /*class ListNode { int val; ListNode next; ListNode(int x) { val = x; } } */ public class Solution { public ListNode reverseList(ListNode head) ...
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. Note: Given m, n satisfy the following condition: ...
public ListNode reverseList(ListNode head) { if (head == null || head.next == null){ return head; } ListNode p = reverseList(head.next); //重点 head.next.next = head; head.next = null; return p; } 总结 三种解法中其实利用栈来反转是最简单的,其次是迭代,最难的是递归,递归可以想成...
publicclassSolution{publicListNodereverseBetween(ListNode head,int m,int n){if(m==n||head==null||head.next==null){returnhead;}ListNode pre=newListNode(0);pre.next=head;ListNode Mpre=pre;ListNode NodeM=head;ListNode NodeN=head;int mNum=1;int nNum=1;while(mNum<m&&NodeM!=null){Mpre=...
王几行xing:【Python-转码刷题】LeetCode 203 移除链表元素 Remove Linked List Elements 提到翻转,熟悉Python的朋友可能马上就想到了列表的 reverse。 1. 读题 2. Python 中的 reverse 方法 list1 = list("abcde") list1.reverse() list1 [Out: ] ['e', 'd', 'c', 'b', 'a'] 2.1 reverse 在...