x):self.val=xself.next=None# 该函数接受原始链表的头节点,返回反转后的链表的头节点defreverseList(head):# 定义两个指针,一个记录当前指标所在的位置,反转的时候,意味着需要把当前节点cur指向它的上一个节点;# 而因为是单链表的缘故,所以当前节点是不知道它的前一个节点是哪一个的,需要一个pre指针去指向...
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given1->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. Analysis: 这道题还是...
Recursion is great to resolve many problems including reverse of a linked list. First let's check the non-recursive version of reverse. To reverse a linked list, we start from the head (in this essay we will use linked list without extra head node). One way is to use a stack, pushes...
(参考视频讲解:Leetcode力扣|206反转链表|递归|reverse linked list_哔哩哔哩_bilibili) # 定义一个链表节点类classListNode:def__init__(self,val=0,next=None):# 初始化函数self.val=val# 节点的值self.next=next# 指向下一个节点的指针# 将给出的数组转换为链表deflinkedlist(list):head=ListNode(list[0]...
A linked list can be reversed either iteratively or recursively. Could you implement both? 题意 将一个链表翻转,有递归和非递归两种方法。 思路 递归 首先判断链表的长度,如果只有一项或为空,直接返回链表即可。 否则,先假装第k+1项及之后所有项都采用递归进行翻转,链表即为n1->n2->……->nk->nk+1-<...
反转链表(Reverse Linked List) 反转一个单链表。如下示例:: 输入: 1->2->3->4->5->NULL 输出: 5->4->3->2->1->NULL public class ListNode { int val; ListNode next; ListNode(int x) { val = x; } } 一、 迭代法: 注意观察示例:1->2->3->4->5->NULL的反转可以看成:NULL<-1<-...
publicListNodereverseList(ListNode head){if(head==null||head.next==null){returnhead;}ListNode n=reverseList(head.next);head.next.next=head;head.next=null;returnn;} 只是注意两个地方: 如果head 是空或者遍历到最后节点的时候,应该返回 head。
Leetcode 92题反转链表 II(Reverse Linked List II) 题目链接 https://leetcode-cn.com/problems/reverse-linked-list-ii/ 题目描述 反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。 说明: 1 ≤ m ≤ n ≤ 链表长度。 示例: 输入: 1->2->3->4->5->NULL, m = 2, n = 4 输出: 1->4...
链接:https://leetcode-cn.com/problems/reverse-linked-list https://leetcode-cn.com/problems/reverse-linked-list/solution/shi-pin-jiang-jie-die-dai-he-di-gui-hen-hswxy/ python # 反转单链表 迭代或递归 class ListNode: def __init__(self, val): ...
}; Python: class Solution(object): def reverseList(self, head): if(head == None or head.next == None): return head newHead = self.reverseList(head.next) head.next.next = head head.next = None return newHead分享至 投诉或建议评论...