What is the inversion method of python linked list

Reversal of python linked list

Reverse linked list

I give you the head node of the singly linked list, please reverse the linked list. And return the reversed linked list.

• Input: head = [1,2,3,4,5]

• Output: [5,4,3,2,1]

• ##Input: head = [1,2]

• Output: [2,1]

Example 3:

• Input ：head = []

• Output: []

Solution

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    """
    解题思路：
    1.新建一个头指针
    2.遍历head链表，依次在新的头节点位置插入，达到反转的效果
    """
    def reverseList(self, head: ListNode) -> ListNode:
        # 循环
        new_head = None

        while head:
            per = head.next # pre 为后置节点，及当前节点的下一个节点

            head.next = new_head # 插入头节点元素

            new_head = head # 把串起来的链表赋值给头指针

            head = per  # 向后移一个单位
        
        return  new_head  # 返回一个新的链表

Python reversal linked list related skills

Given the head node pHead of a singly linked list (the head node has a value, for example, in the figure below, its val is 1), the length is n, after reversing the linked list, return the head of the new linked list.

Requirements: Space complexity O(1)O(1), time complexity O(n)O(n).

Input:

{1,2,3}

Return value:

{3,2,1}

Let’s first look at the most basic reverse linked list code:

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    # 返回ListNode
    def ReverseList(self, pHead):
        # write code here
        cur = pHead
        pre = None
        while cur:
            nextNode = cur.next
            cur.next = pre
            pre = cur
            cur = nextNode
        return pre

Key formula

Catch a few Key points:

• cur: The head node of the original linked list. At the end of the inversion, cur points to the next node of pre

• pre: The tail node of the original linked list is the head node of the reversed linked list. The final return is pre.

• while cur: Indicates the condition for reversing the loop, here is to determine whether cur is empty. You can also change it to other loop conditions according to the conditions of the question

• Reverse the tail node of the linked list. The tail node here is None, and explicit specification will be mentioned later.

For the problem of reversing the linked list, grasp the main roles of the head node of the original linked list, the tail node of the original linked list, the reverse loop condition, and the tail node of the reversed linked list. Basically no problem.

Next, let’s give two examples:

Reversing the specified interval in the linked list

Inverting every k nodes in the linked list

Specified interval reversal in the linked list

Reversing the interval between the m position and the n position of a linked list with a node number of size requires time complexity O(n) and space complexity O(1 ).

Requirements: time complexity O(n), space complexity O(n)

Advanced: time complexity O(n), space complexity O (1)

Input:

{1,2,3,4,5},2,4

Return value:

{1,4,3,2,5}

Apply the formula

The difference between this question and baseline is that Change the reversal of the entire linked list to the reversal of the interval between the m position and the n position of the linked list. Let's apply the formula:

• The head node of the original linked list: cur: starting from head , then take m-1 steps to reach cur

• The tail node of the original linked list: pre: The node in front of cur

• Reverse loop condition : for i in range(n,m)

• Reverse the tail node of the linked list: you need to save it starting from head, then go m-1 steps, when you reach cur, at this time pre position prePos. prePos.next is the tail node of the reverse linked list.

is compared with the previous one. Additional attention is required:

• needs to be saved and started from head. , take m-1 steps again, and when you reach cur, the position of pre at this time is prePos. After the reversal cycle ends, start threading again

• Since the entire linked list is not reversed, it is best to create a new virtual head node dummyNode, and dummyNode.next points to the entire linked list

Code implementation

First look at the code of the formula part:

# 找到pre和cur
i = 1
while i<m:
    pre = cur
    cur = cur.next
    i = i+1
 
# 在指定区间内反转
preHead = pre
while i<=n:
    nextNode = cur.next
    cur.next = pre
    pre = cur
    cur = nextNode
    i = i+1

Threading the needle and thread part of the code:

nextNode = preHead.next
preHead.next = pre
if nextNode:
    nextNode.next = cur

Complete code:

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
 
class Solution:
    def reverseBetween(self , head , m , n ):
        # write code here
        dummpyNode = ListNode(-1)
        dummpyNode.next = head
        pre = dummpyNode
        cur = head
 
        i = 1
        while i<m:
            pre = cur
            cur = cur.next
            i = i+1
 
        preHead = pre
        while i<=n:
            nextNode = cur.next
            cur.next = pre
            pre = cur
            cur = nextNode
            i = i+1
        
        nextNode = preHead.next
        preHead.next = pre
        if nextNode:
            nextNode.next = cur
 
        return dummpyNode.next

Flip the nodes in the linked list every k groups

Flip the nodes in the given linked list every k groups and return the flip The last linked list

If the number of nodes in the linked list is not a multiple of k, keep the last remaining node as is

You cannot change the value in the node, only the node itself.

Requires space complexity O(1), time complexity O(n)

Input:

{1,2, 3,4,5},2

Return value:

{2,1,4,3,5}

Apply formula

The difference between this question and the baseline is that the inversion of the entire linked list is changed to a group of k inversions. If the number of nodes is not a multiple of k, the remaining The nodes remain as they are.

Let’s look at it in sections first. Suppose we face a linked list from position 1 to position k:

• The head node of the original linked list: cur: start from head and go k -1 step to reach cur

• 原链表的尾节点：pre：cur前面的节点

• 反转循环条件：for i in range(1,k)

• 反转链表的尾节点：先定义tail=head，等反转完后tail.next就是反转链表的尾节点

先看下套公式部分的代码：

pre = None
cur = head
tail = head
 
 
i = 1
while i<=k:
    nextNode = cur.next
    cur.next = pre
    pre = cur
    cur = nextNode
    i = i+1

这样，我们就得到了1 位置1-位置k的反转链表。

此时：

• pre：指向反转链表的头节点

• cur：位置k+1的节点，下一段链表的头节点

• tail：反转链表的尾节点

那么，得到位置k+1-位置2k的反转链表，就可以用递归的思路，用tail.next=reverse(cur，k)

需要注意：如果链表中的节点数不是 k 的倍数，将最后剩下的节点保持原样

i = 1
tmp = cur
while i<=k:
    if tmp:
        tmp = tmp.next
    else:
        return head
    i = i+1

代码实现

完整代码：

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
 
class Solution:
    def reverseKGroup(self , head , k ):
       
        # write code here
        return self.reverse(head, k )
    
    def reverse(self , head , k ):
        pre = None
        cur = head
        tail = head
 
        i = 1
        tmp = cur
        while i<=k:
            if tmp:
                tmp = tmp.next
            else:
                return head
            i = i+1
        
        i = 1
        while i<=k:
            nextNode = cur.next
            cur.next = pre
            pre = cur
            cur = nextNode
            i = i+1
 
        tail.next = self.reverse(cur, k)
        return pre

好了，抓住几个关键点：

• cur：原链表的头节点，在反转结束时，cur指向pre的下一个节点

• pre：原链表的尾节点，也就是反转后链表的头节点。最终返回的是pre。

• while cur：表示反转循环的条件，这里是判断cur是否为空。也可以根据题目的条件改成其他循环条件

• 反转链表的尾节点，这里的尾节点是None，后面会提到显式指定。

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