标签：slow else contain over Plan ace return 相等 res
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.
代码：
public class Solution {
public ListNode detectCycle(ListNode head) {
if(head == null || head.next == null) return null;
ListNode slow = head;
ListNode fast = head;
while(slow != null && fast != null){
if(fast.next != null)
fast = fast.next.next;
else
return null;
slow = slow.next;
if(fast == slow) break;
}
if(fast == null)
return null;
int circleNum = 1;
ListNode tmp = slow.next;
while(tmp != slow){
tmp = tmp.next;
circleNum++;
}
fast = head;
slow = head;
for(int i = 0; i < circleNum; i++){
fast = fast.next;
}
while(fast != slow){
fast = fast.next;
slow = slow.next;
}
return slow;
}
}注意：
在第一步找有没有环的时候别丢了相等判断，自己不要犯一些这种小错误
LeetCode 142 链表 Linked List Cycle II
标签：slow else contain over Plan ace return 相等 res
原文地址：https://www.cnblogs.com/muche-moqi/p/12400849.html
