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...
反转链表 II(reverse-linked-list-ii) 秦她的菜 吉利 程序员题目描述 描述 给你单链表的头指针 head 和两个整数 left 和 right ,其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点,返回 反转后的链表。 示例1: 输入:head = [1,2,3,4,5], left = 2, right = 4 输出:...
虚头法 dummy head+递归 Reverse Linked List II 反转链表 Given theheadof a singly linked list and two integersleftandrightwhereleft <= right, reverse the nodes of the list from positionleftto positionright, and returnthe reversed list. 将链表从第left节点开始到第right个节点结束的一段进行翻转 In...
}ListNodep=reverseList(head.next); head.next.next = head; head.next =null;returnp; } } Python 实现 # Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = NoneclassSolution:defreverseList(self, head):""" :type head: ListNode :...
反转链表(Reverse Linked List)C++递归实现 /** * @author:leacoder * @des: 递归实现 反转链表 */class Solution{public:ListNode*reverseList(ListNode*head){//第一个条件是判断递归开始,传入的参数的合法性。第二个是递归的终止条件if(NULL==head||NULL==head->next)returnhead;ListNode*result=reverseList...
/** * Definition for singly-linked list. * type ListNode struct { * Val int * Next *ListNode * } */funcreverseBetween(head*ListNode,leftint,rightint)*ListNode{vhead:=&ListNode{}vhead.Next=headvarf,l,r,t*ListNode f=vheadfori:=1;i<left;i++{f=f.Next}l=f.Next r=l t=r.Next...
In the end, we return the address of the new head node stored in n temporary variable. Consequently, the main program calls the printNodes function for the user to compare modified linked list contents. #include <iostream> #include <string> #include <vector> using std::cin; using std::...
Leetcode 92题反转链表 II(Reverse Linked List II) LeetCode 206题 反转链表(Reverse Linked List) 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-...
kis a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple ofkthen left-out nodes in the end should remain as it is. Example: Given this linked list:1->2->3->4->5 ...
·nt!PsGetNextProcessß Based on the function name here, it appears that we were trying to traverse the linked list of active processes on the system. ·nt!ExpGetProcessInformationß NtQuerySystemInformation() called this function. Based on the function name Get...