链表系列之linked-list-cycle

链表是否有环 I

题目详情

Given a linked list, determine if it has a cycle in it.

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

思路：

• 快慢指针，若快慢指针能相遇则说明链表存在环

代码：

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null)
            return false;
        
        ListNode fast = head;
        ListNode slow = head;
        
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow)
                return true;
        }
        
        return false;
    }
}

链表是否有环 II

题目详情

Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull.

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

思路：

如上图所示：

• 当快慢指针相遇时，
  • 快指针走过的长度：a + b + c + b
  • 慢指针走过的长度：a + b
• 因此，可以知道：
  • 2 * ( a + b ) = a + b + c + d
• 所以，可以得到： a == c

代码：

/**
 * 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 || head.next == null)
            return null;
        
        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (slow == fast) {
                ListNode slow2 = head;
                while (slow != slow2) {
                    slow = slow.next;
                    slow2 = slow2.next;
                }
                return slow;
            }
        }
        return null;
    }
}