Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Note: Do not modify the linked list.
Follow up:
Can you solve it without using extra space?
一刷
方法1： hashMap, 但是空间复杂度为O(n)
方法2：
fast: 2N = D+K+C(i)
slow: N = D+K+C(j)
那么D+K = C(i-2j)
所以先找到fast和slow的相遇点。然后fast从head开始，fast和slow相遇点即为D
因为此时fast走了D + C(i), slow走了C-K + jC
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode detectCycle(ListNode head) {
ListNode slow = head, fast = head;
while(fast!=null && fast.next!=null){
slow = slow.next;
fast = fast.next.next;
if(slow == fast){
fast = head;
while(fast!=slow){
fast = fast.next;
slow = slow.next;
}
return slow;
}
}
return null;
}
}
