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,所以我们需要知道p2的前驱节点p1。 我们首先遍历得到第m -...
http://www.lintcode.com/en/problem/reverse-linked-list-ii/ 【题目解析】 反转整个链表的变种,指定了起点和终点。由于m=1时会变动头节点,所以加入一个dummy头节点 1. 找到原链表中第m-1个节点start:反转后的部分将接回改节点后。 2. 将从p = start->next开始,长度为L = n-m+1的部分链表反转。 3...
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...
reverse(pre, end);returnh.next; }publicvoidreverse(ListNode pre, ListNode end){//B keep to being inserted after pre pre-1-2-3-end, 1= A, 2=B, 3=tListNode A = pre.next, B = A.next;//A 定义为B前面一位, 不断的把B塞到pre后面(也就是list的第一位,然后把原来B的前面A链接到B...
class Solution: # 翻转一个子链表,并且返回新的头与尾 def reverse(self, head: ListNode, tail: ListNode): prev = tail.next p = head while prev != tail: nex = p.next p.next = prev prev = p p = nex return tail, head def reverseKGroup(self, head: ListNode, k: int) -> ListNode...
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,
最近有朋友问我怎么没有更新文章了,因为最近有空的时候都在刷 LeetCode,零零星星刷了快 2 个月了,也累积了不少题目了,所以最近打算把做的几百道题归类,总结一下。所有题目的代码在github.com/halfrost/Le…,每道题都有测试用例和测试代码。 Linked List 的 Tips: ...
個人範例程式碼 – 同向 two pointer,使用 reverse LinkedList 空間 O(1) # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def isPalindrome(self, head: Optional[ListNode]) -> bool: if...
Test Data: Check a stack (using linked list) is empty or not! Stack is empty! Input some elements onto the stack: Stack elements are: 0 1 3 5 6 Size of the stack is 5Sample Solution: C++ Code:#include <iostream> using namespace std; // Define the node structure for the linked...
1 <= m <= n <= length of list. Solution: in-place & one-pass. 1 class Solution { 2 public: 3 ListNode *reverseBetween(ListNode *head, int m, int n) { 4 ListNode dummy(0); 5 ListNode* pre = &dummy; 6 dummy.next = head; ...