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【LeetCode】打卡–Python3算法19. 删除链表的倒数第N个节点
题目
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。
进阶:
你能尝试使用一趟扫描实现吗?
结果–1.暴力法两层循环
执行用时 : 84 ms, 在Remove Nth Node From End of List的Python3提交中击败了21.13% 的用户
内存消耗 : 13.2 MB, 在Remove Nth Node From End of List的Python3提交中击败了33.09% 的用户
Python解答–1.暴力法两层循环
首先使用暴力法两次循环如下
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
N = 0
left = head
while(left):
N = N + 1
left = left.next
if(N==1):
return []
if(N==n):
return head.next
right = head
i = 1
while(i < N-n):
right = right.next
i = i + 1
right.next = right.next.next
return head
结果–2.进阶版单层循环
执行用时 : 56 ms, 在Remove Nth Node From End of List的Python3提交中击败了76.84% 的用户
内存消耗 : 13.2 MB, 在Remove Nth Node From End of List的Python3提交中击败了36.83% 的用户
Python解答–2.进阶版单层循环
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
left = head
right = head
i = 0
while(right.next):
if(i < n):
i = i + 1
else:
left = left.next
right = right.next
if(i == 0 or i < n):
return head.next
elif(n == 1):
left.next = right.next
else:
left.next = left.next.next
return head
我们下次再见,如果还有下次的话!!!
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