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
浮灯Foden24 分钟前
算法-每日一题(DAY13)两数之和
开发语言·数据结构·c++·算法·leetcode·面试·散列表
西工程小巴1 小时前
实践笔记-VSCode与IDE同步问题解决指南;程序总是进入中断服务程序。
c语言·算法·嵌入式
Tina学编程1 小时前
48Days-Day19 | ISBN号,kotori和迷宫,矩阵最长递增路径
java·算法
Moonbit1 小时前
MoonBit Perals Vol.06: MoonBit 与 LLVM 共舞 (上):编译前端实现
后端·算法·编程语言
Nuyoah11klay2 小时前
华清远见25072班C语言学习day11
c语言·指针·回调函数
执子手 吹散苍茫茫烟波2 小时前
leetcode415. 字符串相加
java·leetcode·字符串
百度Geek说3 小时前
第一!百度智能云领跑视觉大模型赛道
算法
big_eleven3 小时前
轻松掌握数据结构:二叉树
后端·算法·面试
big_eleven3 小时前
轻松掌握数据结构:二叉查找树
后端·算法·面试
CoovallyAIHub3 小时前
农田扫描提速37%!基于检测置信度的无人机“智能抽查”路径规划,Coovally一键加速模型落地
深度学习·算法·计算机视觉