141. Linked List Cycle(Linked List-Easy)

最新推荐文章于 2020-09-28 21:09:35 发布

Jack-Cui

最新推荐文章于 2020-09-28 21:09:35 发布

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分类专栏： LeetCode 文章标签： LeetCode 链表 141 Python C-C++

本文链接：https://blog.youkuaiyun.com/c406495762/article/details/55054315

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本文介绍了一种使用快慢指针的方法来判断链表是否存在环。通过一个快指针每次移动两步，慢指针每次移动一步，如果两者相遇则表明链表成环。

转载请注明作者和出处： http://blog.youkuaiyun.com/c406495762

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

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

题目：判断一个给定的列表，是否成环，也就是，判断是否首尾相连。

思路：创建一个快指针，一个慢指针。快指针一次走两布，慢指针一次走一步。慢指针走一圈，快指针走两圈，如果两个指针指向位置相同了，说明列表是环状的。

Language：c

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
bool hasCycle(struct ListNode *head) {
    struct ListNode *fast = (struct ListNode *)malloc(sizeof(struct ListNode));
    struct ListNode *slow = (struct ListNode *)malloc(sizeof(struct ListNode));
    fast = head;
    slow = head;
    while(fast != NULL && fast->next != NULL){
        slow = slow->next;
        fast = fast->next->next;
        if(fast == slow){
            return true;
        }
    }
    return false;
}

Language : cpp

/**
 * 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) {
        ListNode *fast = head;
        ListNode *slow = head;
        while(fast != NULL && fast->next != NULL){
            slow = slow->next;
            fast = fast->next->next;
            if(fast == slow){
                return true;
            }
        }
        return false;
    }
};

Language：python

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

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        try:
            slow = head
            fast = head.next
            while slow is not fast:
                slow = slow.next
                fast = fast.next.next
            return True
        except:
            return False