Reverse a linked list. 翻转一个链表 【题目链接】 http://www.lintcode.com/en/problem/reverse-linked-list/ 【题目解析】 这题要求我们翻转[m, n]区间之间的链表。对于链表翻转来说,几乎都是通用的做法,譬如p1 -> p2 -> p3 -> p4,如果我们要翻转p2和p3,其实就是将p3挂载到p1的后面,所以我们需要知...
/*** Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * }*/publicclassSolution {publicListNode reverseList(ListNode head) { ListNode prev=null; ListNode curr=head;while(curr !=null) { ListNode temp=curr.next; ...
因为tail.next最后转置了。 /*** Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * }*/publicclassSolution {publicListNode reverseBetween(ListNode head,intm,intn) { ListNode dummy=newListNode...
"""classSolution:""" @param head: The first node of the linked list. @return: You should return the head of the reversed linked list. Reverse it in-place. """defreverse(self,head):# case1: empty listifheadisNone:returnhead# case2: only one element listifhead.nextisNone:returnhead#...
Reverse a singly linked list. 翻转一个链表 #1 第一种方法:迭代 代码语言:javascript 代码运行次数:0 classListNode(object):def__init__(self,x):self.val=x self.next=NoneclassSolution(object):defreverseList(self,head):""":type head:ListNode:rtype:ListNode""" ...
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=...
class Solution { public: /** * @param head: The first node of linked list. * @return: The new head of reversed linked list. */ ListNode *reverse(ListNode *head) { // write your code here if(head==NULL||head->next==NULL){ ...
Can you solve this real interview question? Reverse Linked List - Given the head of a singly linked list, reverse the list, and return the reversed list. Example 1: [https://assets.leetcode.com/uploads/2021/02/19/rev1ex1.jpg] Input: head = [1,2,3,
Solution Analysis: Look up the entire linked list in a loop, each time a new parent node is created to point to the current node, and the last generated parent node is the head node after inversion Code: /** * Definition for singly-linked list. ...
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-<...