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  • (链表 双指针) leetcode 142. Linked List Cycle II

    Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

    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.

    Note: Do not modify the linked list.

    Example 1:

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

    Example 3:

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

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

    --------------------------------------------------------------------------------------------------------------------------------------------------------------------------

    这个题是不仅是判断链表中是否存在环,还要返回环的开始的位置。

    用双指针,与141. Linked List Cycle 的类似,关于怎么返回这个位置,可以参考这个大佬的博客:http://www.cnblogs.com/hiddenfox/p/3408931.html

    C++代码:

    /**
     * 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,*fast;
            slow = head;
            fast = head;
            while(true){
                if(fast == NULL || fast->next == NULL)
                    return NULL;
                slow = slow->next;
                fast = fast->next->next;
                if(slow == fast)
                    break;
            }
            slow = head;
            while(slow != fast){
                slow = slow->next;
                fast = fast->next;
            }
            return slow;
        }
    };
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  • 原文地址:https://www.cnblogs.com/Weixu-Liu/p/10704334.html
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