141. Linked List Cycle
給定一個鏈表的頭節點 head，判斷該鏈表中是否存在環。
如果在鏈表中存在某個節點，可以通過不斷地跟隨下一個指針再次到達該節點，則鏈表中存在環。在內部，pos 用於表示尾部的下一個指針連接到的節點的索引。注意，pos 並不作為參數傳遞。
如果鏈表中存在環，返回 true。否則，返回 false。
範例 ：
輸入：head = [3,2,0,-4], pos = 1 輸出：true 解釋：鏈表中存在一個環，其中尾部連接到第 1 個節點（0 索引）。
Python
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
from typing import Optional
class Solution:
def hasCycle(self, head: Optional[ListNode]) -> bool:
if not head or not head.next:
return False
slow = head
fast = head.next
while slow != fast:
if not fast or not fast.next:
return False
slow = slow.next
fast = fast.next.next
return True
18.98MB, 47ms
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
bool hasCycle(ListNode *head) {
if (head == nullptr || head->next == nullptr) {
return false;
}
ListNode* slow = head;
ListNode* fast = head->next;
while (slow != fast) {
if (fast == nullptr || fast->next == nullptr) {
return false;
}
slow = slow->next;
fast = fast->next->next;
}
return true;
}
};
10.47MB, 12ms
Javascript
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {boolean}
*/
var hasCycle = function(head) {
if (head === null || head.next === null) {
return false;
}
let slow = head;
let fast = head.next;
while (slow !== fast) {
if (fast === null || fast.next === null) {
return false;
}
slow = slow.next;
fast = fast.next.next;
}
return true;
};
52.92MB, 66ms
