LeetCode - 142. Linked List Cycle II

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?

快慢指针，参考LeetCode - 141. Linked List Cycle

先判断是否有环，记录第一次相遇位置，然后快指针指向head指针，同步走，再次相遇时的节点即为环的入口。

证明：TODO

/**
 * 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 fast = head, slow = head;
        int cnt = 0;
        while (true) {
            if (fast == null)
                return null;
            if (cnt > 0 && fast == slow)
                break;
            slow = slow.next;
            fast = fast.next;
            if (slow == null || fast == null)
                return null;
            fast = fast.next;
            cnt ++;
        }
        fast = head;
        while (fast != slow) {
            fast = fast.next;
            slow = slow.next;
        }
        return fast;
    }
}