Leetcode 19. 删除链表的倒数第N个节点 ----python

1. 题目描述

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.

2.解题思路

这题要分情况讨论
（1）当链表为空或只有一个节点时，返回结果为None
（2）当删除的元素为链表首元素时，返回head.next
（3）其他情况，定义快慢指针（fast,slow）,快指针先走n步，然后快慢指针一起走直到fast指向为 空，然后删除show指向的节点。

3. 代码实现

class Solution:
    def linkLength(self, root):
        length = 0
        r = root
        while (r):
            length = length + 1
            r = r.next
        return length

    def removeNthFromEnd(self, head, n):
        if(head == None or head.next == None):
            return None
        if(self.linkLength(head) == n):
            return head.next
        root1 = head
        fast = head
        slow = head
        for i in range(0, n): #fast指针向后移动n个元素
            fast = fast.next

        while(fast):#快慢指针同时移动
            beforeSlow = slow
            slow = slow.next
            fast = fast.next #fast指针移到链表的空指针的位置
        beforeSlow.next = beforeSlow.next.next #执行删除操作
        return root1

4. 测试用例及测试结果

测试用例：

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

class Solution:
    def linkLength(self, root):
        length = 0
        r = root
        while (r):
            length = length + 1
            r = r.next
        return length

    def removeNthFromEnd(self, head, n):
        if(head == None or head.next == None):
            return None
        if(self.linkLength(head) == n):
            return head.next
        root1 = head
        fast = head
        slow = head
        for i in range(0, n): #fast指针向后移动n个元素
            fast = fast.next

        while(fast):#快慢指针同时移动
            beforeSlow = slow
            slow = slow.next
            fast = fast.next #fast指针移到链表的空指针的位置
        # print('beforeSlow',beforeSlow.val)
        # print('slow',slow.val)
        beforeSlow.next = beforeSlow.next.next #执行删除操作
        return root1

    def printLink(self,root):
        r = root
        while(r):
            print(r.val)
            r = r.next

root = ListNode(1)
n2 = ListNode(2)
n3 = ListNode(3)
n4 = ListNode(4)
n5 = ListNode(5)

root.next = n2
n2.next = n3
n3.next = n4
n4.next = n5

s = Solution()
result = s.removeNthFromEnd(root,3)
# print(result)
s.printLink(result)

测试结果：

1
2
4
5