移除鏈表倒數第n個元素
Remove Nth Node From End of List
- 給定一個鏈表,移除倒數第n個元素,返回鏈表頭部。
- Given a linked list, remove the nth node from the end of list and return its head.
Note:
Given n will always be valid.
Try to do this in one pass.
example 1
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
example 2
Given linked list: 1, and n = 1.
output: None
example 3
Given linked list: 1->2->3, and n = 3.
output: 2->3
思路
- 兩個指針,
fast
和slow
,fast
指向slow
之後n個位置,同步移動fast
和slow
,當fast.next
爲null的時候,slow.next
即爲要移除的那個元素,只需要slow.next = slow.next.next
即可,時間複雜度O(n) - 注意考慮
n
爲鏈表長度的情況,即移除首個元素
代碼
# Definition for singly-linked list.
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
class Solution(object):
def removeNthFromEnd(self, head, n):
"""
:type head: ListNode
:type n: int
:rtype: ListNode
"""
a = b = head
for i in range(n):
b = b.next
if not b:
return head.next
while b.next:
a = a.next
b = b.next
a.next = a.next.next
return head
本題以及其它leetcode題目代碼github地址: github地址