01_LinkedList

Posted 2022-11-24 pankypan

Linked List

Reverse Linked List

link

Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

Solutions:

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


class Solution:
    def reverse_list(self, head: ListNode) -> ListNode:
        print(self)
        # prev = None
        # cur = head
        prev, cur = None, head
        while cur:
            # temp = cur.next
            # cur.next = prev
            # prev = cur
            # cur = temp
            cur.next, prev, cur = prev, cur, cur.next
        return prev

    def reverse_list_II(self, head:ListNode) -> ListNode:
        """Use recursion"""
        if not head or not head.next:
            return head
        N = self.reverse_list_II(head.next)
        head.next.next = head
        head.next = None
        return N

Swap Nodes in Pairs

link

Given a linked list, swap every two adjacent nodes and return its head.

You may not modify the values in the list‘s nodes, only nodes itself may be changed.

Example:

Given 1->2->3->4, you should return the list as 2->1->4->3.

Solutions:

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


class Solution:
    def swap_pairs(self, head: ListNode) -> ListNode:
        # create a ListNode to record the head
        record_node = ListNode(self)
        record_node.next = head

        # set a cursor
        cur = record_node
        while cur.next and cur.next.next:
            # set two adjoin cursor which attach to the cur cursor
            a = cur.next
            b = a.next

            # change the node.next
            """
            cur.next = b
            a.next = b.next
            b.next = a
            """
            cur.next, a.next, b.next = b, b.next, a

            # move the cur cursor
            cur = a
        return record_node.next

Linked List Cycle

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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?

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


class Solution(object):
    def has_cycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        # set two cursor--a fast && a slow
        cur_fast, cur_slow = head, head

        while cur_slow and cur_fast:
            cur_slow = cur_slow.next
            try:
                cur_fast = cur_fast.next.next
            except AttributeError:
                return False
            
            # the fast cursor meet the slow cursor
            if cur_fast == cur_slow:
                return True
        return False

Linked List Cycle II

link

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?

Solutions:

Consider the following linked list, where E is the cycle entry and X, the crossing point of fast and slow.
        H: distance from head to cycle entry E
        D: distance from E to X
        L: cycle length
        fast: crossing 2 nodes each time
        slow: crossing 1 node each time

                          _____
                         /             head_____H______E                               \\       /
                         X_____/   
        
    
        If fast and slow both start at head, when fast catches slow, slow has traveled H+D and fast 2(H+D). 
        Assume fast has traveled n loops in the cycle, we have:
        2H + 2D = H + D + nL  -->  H + D = nL  --> H = nL - D
        Thus if two pointers start from head and X with same speed, respectively, one first reaches E, the other also reaches E. 

Reverse Nodes in K-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list‘s nodes, only nodes itself may be changed.