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English Version

题目描述

给定一个有环链表,实现一个算法返回环路的开头节点。
有环链表的定义:在链表中某个节点的next元素指向在它前面出现过的节点,则表明该链表存在环路。


示例 1:

输入:head = [3,2,0,-4], pos = 1
输出:tail connects to node index 1
解释:链表中有一个环,其尾部连接到第二个节点。


示例 2:

输入:head = [1,2], pos = 0
输出:tail connects to node index 0
解释:链表中有一个环,其尾部连接到第一个节点。


示例 3:

输入:head = [1], pos = -1
输出:no cycle
解释:链表中没有环。


进阶:
你是否可以不用额外空间解决此题?

解法

先利用快慢指针判断链表是否有环,没有环则直接返回 null

若链表有环,我们分析快慢相遇时走过的距离。

对于慢指针(每次走 1 步),走过的距离为 S=X+Y ①;快指针(每次走 2 步)走过的距离为 2S=X+Y+N(Y+Z) ②。如下图所示,其中 N 表示快指针与慢指针相遇时在环中所走过的圈数,而我们要求的环入口,也即是 X 的距离:

我们根据式子 ①②,得出 X+Y=N(Y+Z) => X=(N-1)(Y+Z)+Z

N=1(快指针在环中走了一圈与慢指针相遇) 时,X=(1-1)(Y+Z)+Z,即 X=Z。此时只要定义一个 p 指针指向头节点,然后慢指针与 p 开始同时走,当慢指针与 p 相遇时,也就到达了环入口,直接返回 p 即可。

N>1时,也是同样的,说明慢指针除了走 Z 步,还需要绕 N-1 圈才能与 p 相遇。

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None


class Solution:
    def detectCycle(self, head: ListNode) -> ListNode:
        slow = fast = head
        has_cycle = False
        while not has_cycle and fast and fast.next:
            slow, fast = slow.next, fast.next.next
            has_cycle = slow == fast
        if not has_cycle:
            return None
        p = head
        while p != slow:
            p, slow = p.next, slow.next
        return p

Java

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode slow = head, fast = head;
        boolean hasCycle = false;
        while (!hasCycle && fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            hasCycle = slow == fast;
        }
        if (!hasCycle) {
            return null;
        }
        ListNode p = head;
        while (p != slow) {
            p = p.next;
            slow = slow.next;
        }
        return p;
    }
}

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* detectCycle(ListNode* head) {
        ListNode* slow = head;
        ListNode* fast = head;
        bool hasCycle = false;
        while (!hasCycle && fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
            hasCycle = slow == fast;
        }
        if (!hasCycle) {
            return nullptr;
        }
        ListNode* p = head;
        while (p != slow) {
            p = p->next;
            slow = slow->next;
        }
        return p;
    }
};

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var detectCycle = function (head) {
    let slow = head;
    let fast = head;
    let hasCycle = false;
    while (!hasCycle && fast && fast.next) {
        slow = slow.next;
        fast = fast.next.next;
        hasCycle = slow == fast;
    }
    if (!hasCycle) {
        return null;
    }
    let p = head;
    while (p != slow) {
        p = p.next;
        slow = slow.next;
    }
    return p;
};

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func detectCycle(head *ListNode) *ListNode {
	slow, fast := head, head
	hasCycle := false
	for !hasCycle && fast != nil && fast.Next != nil {
		slow, fast = slow.Next, fast.Next.Next
		hasCycle = slow == fast
	}
	if !hasCycle {
		return nil
	}
	p := head
	for p != slow {
		p, slow = p.Next, slow.Next
	}
	return p
}

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