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
JieE2121 天前
LeetCode 56. 合并区间|超清晰 JS 图解思路,面试高频区间题
javascript·算法·面试
Jack202 天前
HarmonyOS开发中错误处理策略:网络异常统一处理
算法
小小杨树2 天前
读懂色彩:拍照调色不再难
算法·计算机视觉·配色
JieE2122 天前
LeetCode 226. 翻转二叉树|JS 递归超详细拆解,二叉树入门经典题
javascript·算法
JieE2122 天前
LeetCode 104. 二叉树的最大深度|递归思路超详细拆解
javascript·算法
vivo互联网技术3 天前
CVPR 2026 | 全新强化学习框架 BeautyGRPO:重塑真实人像
算法·大模型·cvpr·影像
Darling噜啦啦3 天前
列表转树算法深度解析:从 Map 到 Reduce 的两种实现,面试高频考点
数据结构·算法·面试
用户497863050733 天前
(一)小红的数组操作
算法·编程语言
怕浪猫3 天前
Electron 系列文章封面图
算法·架构·前端框架