142 Linked List Cycle II

最新推荐文章于 2021-04-07 21:12:48 发布

weixin_39145266

最新推荐文章于 2021-04-07 21:12:48 发布

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分类专栏： LeetCode # Medium # LinkedList # TwoPointers # Attempted

本文链接：https://blog.youkuaiyun.com/weixin_39145266/article/details/88894032

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本文介绍了一种高效检测链表中是否存在循环的方法，并提供了详细的算法实现。通过快慢指针技术，不仅能够判断链表是否有环，还能定位到环的起始节点。

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1 题目

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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: 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.

Follow up:
Can you solve it without using extra space?

2 尝试解

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode* slow = head;
        ListNode* fast = head;
        while(fast != NULL){
            if(!fast->next || !fast->next->next){
                return NULL;                
            }
            fast = fast->next->next;
            slow = slow->next;
            if(slow == fast) break;
        }
        fast = head;
        while(fast!=slow){
            slow = slow->next;
            fast = fast->next;
        }
        return fast;
    }
};

3 标准解

ListNode *detectCycle(ListNode *head) {
    if (head == NULL || head->next == NULL)
        return NULL;
    
    ListNode *slow  = head;
    ListNode *fast  = head;
    ListNode *entry = head;
    
    while (fast->next && fast->next->next) {
        slow = slow->next;
        fast = fast->next->next;
        if (slow == fast) {                      // there is a cycle
            while(slow != entry) {               // found the entry location
                slow  = slow->next;
                entry = entry->next;
            }
            return entry;
        }
    }
    return NULL;                                 // there has no cycle
}