LeetCode 141. Linked List Cycle–百度面试编程题–C++,Python解法

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LeetCode题解专栏：LeetCode题解
我做的所有的LeetCode的题目都放在这个专栏里，大部分题目C++和Python的解法都有。

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题目地址：Linked List Cycle - LeetCode

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Given a linked list, determine if it has a cycle in it.

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.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
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: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

图见题目网址

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

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这道题目是早上百度现场面试的编程题。
我当时想到的解法是用set来存所有节点，如果遍历到已经存在的节点，那么这个链表存在环。
Python解法如下：

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

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        s=set()
        while head!=None:
            if head in s:
                return True
            s.add(head)
            head=head.next
        return False

时间复杂度和空间复杂度都是O(n)。
面试官问我有其他解法吗，我想不出来。。。。。。。。。。。。。

空间复杂度O(1)的做法是使用快慢指针

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

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        if head is None:
            return False
        slow, fast = head, head.next
        while fast is not None and fast.next is not None:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                return True
        return False

C++解法如下：

/**
 * 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;
    }
};