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HomeBackend DevelopmentPython TutorialWhat 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.

    What is the inversion method of python linked list

    • Input: head = [1,2,3,4,5]

    • Output: [5,4,3,2,1]

    What is the inversion method of python linked list

    • ##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).

    What is the inversion method of python linked list

    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

    What is the inversion method of python 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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