• 142. Linked List Cycle II


    Problem:

    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.
    

    思路
    和141题类似的思路,也是设置fast和slow2个指针,fast每次移动2个节点,slow每次移动一个节点。设链表开始节点为start,环开始的节点为entry,fast和slow相遇的节点为meet。从start到entry距离为(x_1),从entry到meet的距离为(x_2),然后从meet沿环到entry的距离为(x_3)。则fast指针移动的距离为(x_1+x_2+x_3+x_2),slow指针移动的距离为(x_1+x_2),可以得出等式:(x_1+x_2+x_3+x_2 = 2(x_1+x_2)),可以推出(x_1=x_3)。也就是说,若设置一个新的节点to_entry从start开始移动,slow从meet开始移动,2者移动相同的距离就会在entry节点相遇,而此时移动的节点数也就是entry的下标,即找到了环开始的节点。

    Solution:

    ListNode *detectCycle(ListNode *head) {
        if (!head)  return NULL;
        
        ListNode *fast = head, *slow = head, *entry = head;
        
        while (fast->next && fast->next->next) {
            fast = fast->next->next;
            slow = slow->next;
            if (fast == slow) {
                while (slow != entry) {
                    slow = slow->next;
                    entry = entry->next;
                }
                return entry;
            }
        }
        return NULL;
    }
    

    性能
    Runtime: 12 ms  Memory Usage: 9.9 MB

    相关链接如下:

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  • 原文地址:https://www.cnblogs.com/dysjtu1995/p/12274503.html
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