Leetcode - 142 - Linked List Cycle II

本文介绍了一种使用快慢指针的方法来解决链表中寻找循环起点的问题,并提供了C++、Java和Python三种语言的实现代码。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Given 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. Note that pos is not passed as a parameter.

Notice that you should not modify the linked list.

 

 

Solutions: Fast and Slow Pointers

The core of the problem is to find that

distance(the meeting point, the first node of the cycle) = distance( the first node of the list, the first node of the cycle).

It can be proved by:

x3 = x1 meas two pointers start from 'start' and 'meet' seperately with the same spped can meet at 'cycle'.

 

C++:

ListNode *detectCycle(ListNode *head) {
    if (head == NULL || head->next == NULL) return NULL;
    
    ListNode* firstp = head;
    ListNode* secondp = head;
    bool isCycle = false;
    
    while(firstp != NULL && secondp != NULL) {
        firstp = firstp->next;
        if (secondp->next == NULL) return NULL;
        secondp = secondp->next->next;
        if (firstp == secondp) { isCycle = true; break; }
    }
    
    if(!isCycle) return NULL;
    firstp = head;
    while( firstp != secondp) {
        firstp = firstp->next;
        secondp = secondp->next;
    }

    return firstp;
}
// https://leetcode.com/problems/linked-list-cycle-ii/discuss/44793/O(n)-solution-by-using-two-pointers-without-change-anything


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
}
// https://leetcode.com/problems/linked-list-cycle-ii/discuss/44781/Concise-O(n)-solution-by-using-C%2B%2B-with-Detailed-Alogrithm-Description

Java:

public ListNode detectCycle(ListNode head) {
    if (head == null || head.next == null) return null;
    ListNode slow = head, fast = head, start = head;
    while (fast != null && fast.next != null) {
        slow = slow.next;
        fast = fast.next.next;
        if (slow == fast) {
            while (slow != start) {
                slow = slow.next;
                start = start.next;
            }
            return start;
        }
    }
    return null;
}
// by annieqt https://leetcode.com/problems/linked-list-cycle-ii/discuss/44793/O(n)-solution-by-using-two-pointers-without-change-anything

Python:

lass Solution:
    # @param head, a ListNode
    # @return a list node
    def detectCycle(self, head):
        try:
            fast = head.next
            slow = head
            while fast is not slow:
                fast = fast.next.next
                slow = slow.next
        except:
            # if there is an exception, we reach the end and there is no cycle
            return None

        # since fast starts at head.next, we need to move slow one step forward
        slow = slow.next
        while head is not slow:
            head = head.next
            slow = slow.next

        return head
# https://leetcode.com/problems/linked-list-cycle-ii/discuss/44783/Share-my-python-solution-with-detailed-explanation


class Solution:
    # @param head, a ListNode
    # @return a list node
    def detectCycle(self, head):
        if head == None: return None
        hare, turtle= head, head
        while hare != None:
            turtle = turtle.next
            hare = hare.next
            if hare == None: return None
            hare = hare.next
            if hare == turtle:
                turtle = head
                while turtle != hare:
                    hare = hare.next
                    turtle = turtle.next
                return hare
        return None

# https://leetcode.com/problems/linked-list-cycle-ii/discuss/44793/O(n)-solution-by-using-two-pointers-without-change-anything

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值