LeetCode---Linked List Cycle、Linked List Cycle II、Intersection of Two Linked Lists

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最新推荐文章于 2022-08-06 10:20:24 发布

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分类专栏：刷题练习 Linked List

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141. Linked List Cycle

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

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.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
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: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

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

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

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。

快慢指针法。定义两个指针：慢指针每次走一步；快指针每次走两步。依次循环下去，如果链表存在环，那么快慢指针一定会有相等的时候。

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        slow=fast=head
        while fast and fast.next:
            slow=slow.next
            fast=fast.next.next
            if slow==fast:
                return True
        return False

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.

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?

给定一个链表，返回链表开始入环的第一个节点。如果链表无环，则返回 null。

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。

说明：不允许修改给定的链表。

在Linked List Cycle中，通过双指针方法来判断链表中是否存在环。在此基础上，我们来找出环的起始节点。如下图所示，假设链表的起始节点为A，环的起始节点为B，快慢指针在C处相遇。因为快指针的速度是慢指针的两倍，所以在相同时间内，它走过的路程是慢指针的两倍，而快指针走过的路程是(x+y+z+y)，而慢指针走过的路程是(x+y)，根据关系我们可以得到x+y+z+y = 2(x+y)，也就是说x=z。此时快慢指针在C处，头指针在A处，而它们到B的距离相等，那么只要有两个指针分别从点A和点C出以相同的速度前进就会在点B处相遇，也就是找到了环的起始节点。

注：上面的情况是理想情况，实际上在点C相遇时，快指针可能已经绕着环走了好几圈了，如AB很长，而环很小的情况。假设走了n圈，此时等式为x+y+n(z+y) = 2(x+y)，即x=(n-1)(y+z)+z，而从点C绕n-1圈后再走z的距离还是会跟从点A出发的指针在点B相遇。

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def detectCycle(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        slow=fast=head
        while fast and fast.next:
            slow=slow.next
            fast=fast.next.next
            if slow==fast:
                slow=head
                while slow!=fast:
                    slow=slow.next
                    fast=fast.next
                return slow
        return None

160. Intersection of Two Linked Lists

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

begin to intersect at node c1.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Example 2:

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

Notes:

If the two linked lists have no intersection at all, return null.
The linked lists must retain their original structure after the function returns.
You may assume there are no cycles anywhere in the entire linked structure.
Your code should preferably run in O(n) time and use only O(1) memory.

编写一个程序，找到两个单链表相交的起始节点。

如果有一个链表为空，返回NULL；先遍历链表得到A、B的长度，如果最后一个节点不同，返回NULL；使长的链表先走长度差步，然后一起往后走，直到节点相同返回即可。

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        if headA is None or headB is None:
            return None
        lenA=1
        hA=headA
        while hA.next !=None:
            lenA+=1
            hA=hA.next
        lenB=1
        hB=headB
        while hB.next != None:
            lenB+=1
            hB=hB.next
        if hA!=hB:
            return None
        hA=headA
        hB=headB
        if lenA>lenB:
            for i in range(lenA-lenB):
                hA=hA.next
        else:
            for i in range(lenB-lenA):
                hB=hB.next
        while hA!=hB:
            hA=hA.next
            hB=hB.next
        return hA