Leetcode算法系列（链表）之删除链表倒数第N个节点

Posted 2020-11-15 超级英雄拯救世界之前成长的日子

Leetcode算法系列（链表）之删除链表倒数第N个节点

难度：中等
给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点。
示例：
给定一个链表: 1->2->3->4->5, 和 n = 2.
当删除了倒数第二个节点后，链表变为 1->2->3->5.
说明：
给定的 n 保证是有效的。
链接：https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list

Python实现

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


class Solution(object):
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        """
        方法一：
        1. 复制表头元素head1=head
        2. 遍历单链表
            将当前节点复制为head2
            遍历当前节点后n个元素
                若不够n个元素就到达链表末尾，返回整个链表恰好n位，删除首节点即可，即返回首节点下一节点
                head2后移
            若head2后移n个元素之后正好为空，则删除该节点head下一节点
            返回表头元素head1
        """
        i = 0
        head1 = head
        while head:
            head2 = head
            j = 0
            while j < n + 1:
                if not head2:
                    return head1.next
                head2 = head2.next
                j += 1
            if not head2:
                head.next = head.next.next
                return head1
            i += 1
            head = head.next            

    def removeNthFromEnd2(self, head: ListNode, n: int) -> ListNode:
        """
        一次遍历，滑动串口
        """
        fast = head
        slow = head
        for i in range(n):
            fast = fast.next
        if not fast:
            # 链表长度为N的情况
            return head.next

        while fast.next:
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return head

    def removeNthFromEnd3(self, head: ListNode, n: int) -> ListNode:
        dummy = ListNode(0)
        dummy.next = head
        fast,slow = dummy,dummy
        for  i in range(n+1):
            fast = fast.next
        while fast is not None: 
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return dummy.next


def create_listnode(list1: list) -> ListNode:
    print(list1)
    list1_nodes = [ListNode(x=node)  for node in list1]
    i = 0
    while i < len(list1) - 1:
        list1_nodes[i].next = list1_nodes[i + 1]
        i += 1

    return list1_nodes[0]


def print_lnode(lnode):
    while lnode:
        print(lnode.val)
        lnode = lnode.next

if __name__ == "__main__":
    l1 = create_listnode(list1=[1,2,3,4,5])
    solution = Solution()
    # head = solution.removeNthFromEnd(head=l1, n=1)
    head = solution.removeNthFromEnd2(head=l1, n=2)
    # if head.__class__ == ListNode:
    print_lnode(head)

Go语言实现

package main

import "fmt"

// Definition for singly-linked list.
type ListNode struct {
    Val  int
    Next *ListNode
}

func (h *ListNode) Show() {
    fmt.Println(h.Val)
    for h.Next != nil {
        h = h.Next
        fmt.Println(h.Val)
    }
}

func removeNthFromEnd1(head *ListNode, n int) *ListNode {
    // 快慢指针法 先让快指针领先n个位置
    // 当快指针到达nil时 慢指针即倒数第n个位置
    // 0 ms 2.2 MB  Golang
    node := &ListNode{Next: head}
    fast, slow, step := node, node, 0
    for step < n {
        fast = fast.Next
        step++
    }
    for fast.Next != nil {
        fast = fast.Next
        slow = slow.Next
    }
    slow.Next = slow.Next.Next
    return node.Next
}

func removeNthFromEnd2(head *ListNode, n int) *ListNode {
    // 数组循环存储最后N个元素，一次遍历
    // 题目规定“给定的 n 保证是有效的。”所以不对n进行检查了
    //  0 ms    2.2 MB  Golang
    var length int = n + 1
    var tempNodes []*ListNode = make([]*ListNode, length)
    var countNode int = 0
    var tail *ListNode = head
    for tail != nil {
        tempNodes[countNode%length] = tail
        tail = tail.Next
        countNode++
    }
    if countNode == n { // 最后一个节点的情况
        return head.Next
    }
    if n == 1 { // 第一个节点的情况
        tempNodes[countNode%length].Next = nil
    } else { // 中间的情况
        tempNodes[countNode%length].Next = tempNodes[(countNode+2)%length]
    }
    return head
}

func removeNthFromEnd3(head *ListNode, n int) *ListNode {
    //  两次遍历
    // 0 ms 2.2 MB  Golang
    if head.Next == nil {
        return nil
    }
    node := &ListNode{Next: head}
    pointer, pointer2, length := node, node, 1
    for pointer.Next != nil {
        pointer = pointer.Next
        length++
    }
    index := length - n
    for i := 1; i <= length; i++ {
        if i == index {
            if pointer2.Next == nil {
                break
            }
            pointer2.Next = pointer2.Next.Next
            break
        } else {
            pointer2 = pointer2.Next
        }
    }
    return node.Next
}

func removeNthFromEnd4(head *ListNode, n int) *ListNode {
    // 递归实现  0 ms   2.2 MB  Golang
    head, _ = handler(head, 1, n)
    return head
}

func handler(head *ListNode, layer, n int) (*ListNode, int) {
    if head == nil {
        return head, layer - 1
    }
    next, maxNum := handler(head.Next, layer+1, n)
    if layer == maxNum-n+1 {
        return head.Next, maxNum
    } else if layer == maxNum-n {
        head.Next = next
        return head, maxNum
    } else {
        return head, maxNum
    }
}

func create_link_list(list1 []int) *ListNode {
    head := &ListNode{Val: list1[0]}
    tail := head
    for i := 1; i < len(list1); i++ {
        tail.Next = &ListNode{Val: list1[i]}
        tail = tail.Next
        // head.Append(list1)
    }
    return head
}

func main() {
    l1 := []int{1, 2, 3, 4, 5}
    fmt.Println(l1)
    head1 := create_link_list(l1)
    head2 := removeNthFromEnd4(head1, 2)
    head2.Show()
}

 

执行结果

方法	执行用时	内存消耗	语言
python-遍历判断	40 ms	13.9 MB	Python3
python-快慢指针，滑动窗口	40 ms	13.8 MB	Python3
GO-快慢指针，滑动窗口	0ms	2.2MB	Golang
Go-数组存储，一次遍历	0ms	2.2MB	Golang
Go-两次遍历	0ms	2.2MB	Golang
Go-递归实现	0ms	2.2MB	Golang