环形链表

题目

给定一个链表, 判断链表中是否有环, 若有则返回链表尾指向的结点.

解法一

遍历时使用 HashSet 保存所有节点, 若遍历到重复节点则返回该节点

Java

class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {
        val = x;
        next = null;
    }
}

public class CycleList {
    public static ListNode hasCycle(ListNode head) {
        HashSet<ListNode> nodeSet = new HashSet<>();

        while (head != null) {
            if (nodeSet.contains(head))
                return head;
            nodeSet.add(head);
            head = head.next;
        }

        return null;
    }
}

Python

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
        
class CycleList:
    def hasCycle(self, head):     
        node_set = set()
        while head:
            if head in node_set:
                return head
            node_set.add(head)
            head = head.next
        
        return None

解法二

使用快慢指针, 该解法空间复杂度为 O(1) , 慢指针每次移动一个位置, 快指针每次移动两个位置, 该方法利用了两个定律:

• 若链表中有环, 则两指针必相遇
• 两指针相遇后, 快指针放回头结点, 从此快慢指针都每次移动一个位置, 两指针必在第一个入环节点处相遇

Java

class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {
        val = x;
        next = null;
    }
}

public class CycleList {
	public static ListNode hasCycle(ListNode head) {
        if (head == null || head.next == null || head.next.next == null)
            return null;

        ListNode slow = head.next;
        ListNode fast = head.next.next;

        while (slow != fast) {
            if (fast.next == null || fast.next.next ==null)
                return null;
            slow = slow.next;
            fast = fast.next.next;
        }

        fast = head;

        while (fast != slow) {
            slow = slow.next;
            fast = fast.next;
        }

        return slow;
    }
}

Python

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class CycleList:
    def hasCycle(self, head):
        if head and head.next and head.next.next:
            slow = head.next
            fast = head.next.next
            
            while slow != fast:
                if fast.next and fast.next.next:
                    slow = slow.next
                    fast = fast.next.next
                else:
                    return None
            
            fast = head
            while slow != fast:
                slow = slow.next
                fast = fast.next
            
            return fast
        else:
            return None

例题 LeetCode # 141

给定一个链表，判断链表中是否有环。

为了表示给定链表中的环，我们使用整数 pos 来表示链表尾连接到链表中的位置(索引从 0 开始)。 如果 pos 是 -1，则在该链表中没有环。

示例 1：

输入：head = [3,2,0,-4], pos = 1
输出：true
解释：链表中有一个环，其尾部连接到第二个节点。

示例 2：

输入：head = [1,2], pos = 0
输出：true
解释：链表中有一个环，其尾部连接到第一个节点。

示例 3：

输入：head = [1], pos = -1
输出：false
解释：链表中没有环。

要求：

空间复杂度为 O(1)

Java

class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {
        val = x;
        next = null;
    }
}

public class CycleList {
	public static boolean hasCycle(ListNode head) {
        if (head == null || head.next == null || head.next.next == null)
            return false;

        ListNode slow = head.next;
        ListNode fast = head.next.next;

        while (slow != fast) {
            if (fast.next == null || fast.next.next ==null)
                return false;
            slow = slow.next;
            fast = fast.next.next;
        }

        return true;
    }
}

Python

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class CycleList:
	def hasCycle(self, head):
        if head and head.next and head.next.next:
            slow = head.next
            fast = head.next.next
            
            while slow != fast:
                if fast.next and fast.next.next:
                    slow = slow.next
                    fast = fast.next.next
                else:
                    return False
                    
            return True
        else:
            return False