142. Linked List Cycle II

Medium

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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.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

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

1. 暴力解决：

定义一个时间，遍历链表看看能不能发现null。

2. Set判重：

边遍历边将存入set，并将节点与set 中的数据比较。

时间复杂度O（n）；

3. 快慢指针法：

定义两个快慢指针 pFast 和 pSlow，pFast 每次走两步，pSlow 每次走一步，如果 pFast 在循环遍历后为 null，则链表中不存在环。如果 pFast 和 pSlow 相遇则链表中存在环。

/**
 * 检查是否存在环形链表
 *
 * @param head
 * @return
 */
public boolean detectLoop(Node head) {
    Node pSlow = head, pFast = head;
    boolean detectFlag = false;
    // 只包含一个结点
    if (head.next == null) {
        return detectFlag;
    }

    List<Integer> slowPassNodes = new ArrayList<>();
    List<Integer> fastPassNodes = new ArrayList<>();

    while (true) {
        pSlow = pSlow.next;
        pFast = pFast.next.next;
        slowPassNodes.add(pSlow.item);
        if (pFast != null) {
            fastPassNodes.add(pFast.item);
        }
        if (pFast == null) {
            break;
        }
        if (pSlow == pFast) {
            detectFlag = true;
            break;
        }
    }
    System.out.println("slow pointer traverse node list: " + slowPassNodes);
    System.out.println("fast pointer traverse node list: " + fastPassNodes);
    return detectFlag;
}