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
wuminyu15 分钟前
Continuation::freeze与thaw的x86_64汇编Stack Chunk拆解源码分析
java·linux·c语言·汇编·jvm·c++
纪念 22931 分钟前
C++算法(三)
开发语言·c++·算法
学不会pwn不改名1 小时前
时间复杂度与空间复杂度详解
算法
AbandonForce1 小时前
递归、搜索与回溯:从板书到二叉树解题方法
c++·算法·决策树
打工仔折腾 AI1 小时前
让时序预测真正参与决策:TimechoAI 工程化落地与管线设计实践
人工智能·后端·python·算法·机器学习·性能优化
Light601 小时前
企业自建 AI 智能体,成本差 8 倍还是 46 倍?
java·人工智能·算法·知识图谱·rag·ai agent 上下文·token 成本
曹牧1 小时前
VS: Visual Studio 里把未包含的文件加进项目2
linux·运维·算法
重生之小比特2 小时前
【算法拾光】滑动窗口|长度最小的子数组 + 无重复字符的最长子串
算法
weixin_307779132 小时前
C++代码实现MATLAB中的fitnlm函数功能
开发语言·c++·算法·matlab