LeetCode //C - 142. Linked List Cycle II

142. Linked List Cycle II

Given the head of a linked list, return the node where the cycle begins. If there is no cycle, return null.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

Do not modify the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

Constraints:
  • he number of the nodes in the list is in the range [ 0 , 1 0 4 ] [0, 10^4] [0,104].
  • − 1 0 5 < = N o d e . v a l < = 1 0 5 -10^5 <= Node.val <= 10^5 −105<=Node.val<=105
  • pos is -1 or a valid index in the linked-list.

From: LeetCode

Link: 142. Linked List Cycle II


Solution:

Ideas:
  1. Initialization: Start with two pointers at the head of the linked list, slow and fast.

  2. Movement: Move slow by one node and fast by two nodes at each step. The slow pointer moves one step at a time (slow = slow->next;), while the fast pointer moves two steps at a time (fast = fast->next->next;).

  3. Cycle Detection: If there is a cycle, the fast pointer will eventually overlap with the slow pointer inside the cycle since the fast pointer is moving faster. If the fast pointer reaches NULL (i.e., fast == NULL || fast->next == NULL), that means the list has an end and, therefore, no cycle.

  4. Identifying Cycle Entry: Once a cycle is detected (i.e., slow == fast), move the slow pointer back to the head of the list and keep the fast pointer at the meeting point. Now move both pointers at the same pace, one step at a time (slow = slow->next; fast = fast->next;).

  5. Cycle Entry Point: The point where the slow and fast pointers meet again is the start of the cycle. This happens because the distance from the head of the list to the start of the cycle is the same as the distance from the meeting point to the start of the cycle following the cycle's path.

Code:
c 复制代码
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode *detectCycle(struct ListNode *head) {
    struct ListNode *slow = head;
    struct ListNode *fast = head;
    
    // First step: Determine whether there is a cycle in the list.
    while (fast != NULL && fast->next != NULL) {
        slow = slow->next;
        fast = fast->next->next;
        
        if (slow == fast) {
            // Cycle detected, now let's find the entry point.
            slow = head; // Move slow pointer to head.
            while (slow != fast) {
                slow = slow->next;
                fast = fast->next;
            }
            return slow; // slow is now the start of the cycle.
        }
    }
    return NULL; // No cycle found.
}
相关推荐
pianmian12 小时前
python数据结构基础(7)
数据结构·算法
闲晨2 小时前
C++ 继承:代码传承的魔法棒,开启奇幻编程之旅
java·c语言·开发语言·c++·经验分享
好奇龙猫4 小时前
【学习AI-相关路程-mnist手写数字分类-win-硬件:windows-自我学习AI-实验步骤-全连接神经网络(BPnetwork)-操作流程(3) 】
人工智能·算法
sp_fyf_20244 小时前
计算机前沿技术-人工智能算法-大语言模型-最新研究进展-2024-11-01
人工智能·深度学习·神经网络·算法·机器学习·语言模型·数据挖掘
香菜大丸5 小时前
链表的归并排序
数据结构·算法·链表
jrrz08285 小时前
LeetCode 热题100(七)【链表】(1)
数据结构·c++·算法·leetcode·链表
oliveira-time5 小时前
golang学习2
算法
南宫生6 小时前
贪心算法习题其四【力扣】【算法学习day.21】
学习·算法·leetcode·链表·贪心算法
懒惰才能让科技进步6 小时前
从零学习大模型(十二)-----基于梯度的重要性剪枝(Gradient-based Pruning)
人工智能·深度学习·学习·算法·chatgpt·transformer·剪枝
DARLING Zero two♡7 小时前
关于我、重生到500年前凭借C语言改变世界科技vlog.16——万字详解指针概念及技巧
c语言·开发语言·科技