Leetcode Linked List Cycle II

时间:2022-11-29 07:39:11

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

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

借用博客http://www.cnblogs.com/hiddenfox/p/3408931.html的图

Leetcode Linked List Cycle II

设环的距离为L = (b+c),无环的距离为a,

假设在时间t相遇,慢指针行驶距离为x,则 1 x t = x,即t=x

则快指针行驶的距离为2t = 2x,

则慢指针环上停留点为(x-a)%L,快指针停留点为 (2x-a)%L,由于在t时刻相遇,故(x-a)%L = (2x-a)%L,

根据同余定理 x%L = 0,

当快指针相遇后减慢速度为1,快指针从z继续行走a长度停下,则行走的路程为2x+a,在环上的位置为(2x+a-a)%L = 2x%L=2(x%L) = 0,即回到环的起始点,故知道环的起始点

ListNode* hasCycle(ListNode* head){
if(head == NULL || head->next == NULL) return false;
ListNode* first = head, *second = head;
while(second!=NULL && second->next!=NULL){
first = first->next;
second = second->next->next;
if(first == second) return first;
}
return NULL;
} ListNode *detectCycle(ListNode *head){
ListNode* cycleNode = hasCycle(head);
if(cycleNode != NULL){
ListNode* startNode = head;
while(startNode!=cycleNode){
cycleNode = cycleNode->next;
startNode = startNode->next;
}
}
return cycleNode;
}