142. Linked List Cycle II

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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?

解法1：

一个老题。快慢两个指针，等相遇的时候slow继续往前走一个然后继续前行，fast回到原点然后一次一步直到再次相遇就是他们的交点。
C++

1

Java

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/**
 * 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) {
        if (head == null) {
            return null;
        }
        ListNode slow = head, fast = head.next;
        while (fast != slow) {
            if (fast == null || fast.next == null) {
                return null;
            }
            fast = fast.next.next;
            slow = slow.next;
        }
        slow = slow.next;
        fast = head;
        while (fast != slow) {
            fast = fast.next;
            slow = slow.next;
        }
        return fast;
    }
}