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。
进阶:你能尝试使用一趟扫描实现吗?
示例 1:
输入:head = [1,2,3,4,5], n = 2
输出:[1,2,3,5]
示例 2:
输入:head = [1], n = 1
输出:[]
示例 3:
输入:head = [1,2], n = 1
输出:[1]
提示:
链表中结点的数目为 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list
python
# 0019.删除倒数第N个节点
# 参考:https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/solution/dong-hua-tu-jie-leetcode-di-19-hao-wen-ti-shan-chu/
class ListNode:
def __init__(self, val):
self.val = val
self.next = None
class Solution:
def removeNthFromENd(self, head: ListNode, n: int) -> ListNode:
"""
双指针,一趟遍历,时间O(n), 空间O(1)
思路:
- 假设链表长度L(未知), 设置虚拟节点,指向head, 初始化p1,p2指针,从虚拟节点开始
- 固定p1,移动p2指针n+1步
- 同时移动p1,p2,直到p2指针到末尾,
- 此时p2指针又走了L-n-1步, 同理p1指针也走了L-n-1步,p1指针的下个节点即为删除节点,删除即可
- 最后返回head节点
:param head:
:param n:
:return:
"""
dummyHead = ListNode(0)
dummyHead.next = head
p1, p2 = dummyHead, dummyHead
for i in range(0, n+1):
p2 = p2.next
while p2:
p1 = p1.next
p2 = p2.next
p1.next = p1.next.next
retNode = dummyHead.next
dummyHead.next = None
return retNode
golang
// 双指针
func removeNthFromEnd(head *ListNode, n int) *ListNode {
dummyHead := &ListNode{}
dummyHead.Next = head
p1, p2 := dummyHead, dummyHead
for i := 0; i < n+1; i++ {
p2 = p2.Next
}
for p2 != nil {
p1 = p1.Next
p2 = p2.Next
}
p1.Next = p1.Next.Next
retNode := dummyHead.Next
dummyHead.Next = nil
return retNode
}