slow 和 fast 节点分别初始化为节点 1 和 2,假设快慢指针第一次相遇的节点为 0 对应于环中的第 i 个节点 Ci, 那么此时慢指针正好走了 n−r−1+i 步,快指针则走了 2⋅(n−r−1+i) 步,且存在式1: n−r−1+i+1=l⋅r. (之所以在 i 后面加 1 是因为快指针初始化时多走了一步)...
if(head == NULL || head->next==NULL || head->next->next==NULL) return NULL; ListNode* p=head; ListNode* q=head; while(1) { if(p->next == NULL) return NULL; p = p->next; q = q->next; if(p->next != NULL) { p = p->next; if(p == q) { break; } } } q =...
public ListNode detectCycle(ListNode head) { HashSet<ListNode> set = new HashSet<>(); while (head != null) { set.add(head); head = head.next; if (set.contains(head)) { return head; } } return null; } 解法二 快慢指针 还是之前的思想, 学数据结构课程的时候,应该都用过这个方法,很...
*/publicclassSolution{publicListNodedetectCycle(ListNode head){if(head==null||head.next==null)returnnull;ListNode runner=head;ListNode walker=head;while(runner.next!=null&&runner.next.next!=null){walker=walker.next;runner=runner.next.next;if(walker==runner){ListNode walker2=head;while(walker!=wa...
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. 翻译过来就是找到链表中的环,并返回环起点的节点。
Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull. Follow up: Can you solve it without using extra space? 快慢指针找重合判断是否循环、而后fast从头开始+1,slow继续前进直到重合 思路解析: 首先我们看一张图; ...
Example 2: Input:head=[1,2],pos=0Output:tail connects to node index0Explanation:There is a cycle in the linked list,where tail connects to the first node. 1. 2. 3. Example 3: Input:head=[1],pos=-1Output:no cycle Explanation:There is no cycle in the linked list. ...
Given a linked list, return the node where the cycle begins. If there is no cycle, return null. 译文:给定一个链表,如果链表中存在环,则返回到链表中环的起始节点,如果没有环,返回null。 知识点 Linked List Cycle I —— 通过快慢两个指针来确定链表上是否存在环 快慢指针的相遇点到环入口点的距离 ...
if(slow==fast)returnfast;[1,2]tail connects to node index0 第一次检查cycle时候 fast == slow == 1 接下来循环入口如果不判断,之后两个会在[2]判断相等并返回[2]作为cycle的头结点,明显是不对的。 structListNode*detectCycle(structListNode*head){if(head==NULL)returnNULL;structListNode*slow,*fast...
If any of the conditions are not true, then the switch is not in the affected list and no action is required. Serial Number Validation This field notice provides the ability to determine if the serial number(s) of a device is impacted by this issue. In order to verify your serial number...