描述
给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。
思路
双指针技巧,快慢指针。
Code
- 提交Code
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
fast = head
slow = head
while(n >= 0):
if(fast == None):
head = head.next
return head
fast = fast.next
n -= 1
while( fast != None):
fast = fast.next
slow = slow.next
slow.next = slow.next.next
return head
- 完整Code
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
def removeNthFromEnd(self, head:ListNode,n:int):
listN = ListNode(head[0])
tmp = listN
# 环状链表的构造
for m in range(1,head.__len__()):
listN.next = ListNode(head[m])
listN = listN.next
print('-----------------------')
fast = tmp
slow = tmp
# if(n == 5):
# tmp = tmp.next
# return tmp
while (n >= 0):
# 用于判断是否为移除【正数】第一个数字的情况
if(fast == None):
tmp = tmp.next
return tmp
fast = fast.next
n -= 1
print('888888888888888888888888888888')
while (fast != None):
fast = fast.next
slow = slow.next
slow.next = slow.next.next
print('888888888888888888888888888888')
return tmp
head = [1,2]
S = Solution()
print("###################result###################")
result = S.removeNthFromEnd(head,1)
print("result"+str(result))
print('-------------------------------------------')
while result != None:
print(result.val)
result = result.next
