这个题目的意思是删除从尾部起第N个节点。
Loading...leetcode.comGiven a linked list, remove then-th node from the end of list and return its head.
Example:
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.
这个题目的解题技巧，设置二个快慢指针，让快指针先走N步后，慢指针和快指针同时往后移动，当快指针移动到尾部后，这个时候慢指针的下一个节点就是要被删除的节点。
请看Python 代码 和Go 代码
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
// 黄哥Python培训 黄哥所写
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
if head is None:
return None
fast, slow = head, head
for _ in range(n):
fast = fast.next
if fast is None:
head = head.next
return head
while fast.next != None:
fast = fast.next
slow = slow.next
slow.next = slow.next.next
return head下面是Go 语言代码
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func removeNthFromEnd(head *ListNode, n int) *ListNode {
var fast, slow *ListNode
fast = head
slow = head
for i := 0; i < n; i++ {
fast=fast.Next
}
if fast == nil {
head = head.Next
return head
}
for fast.Next != nil {
fast = fast.Next
slow = slow.Next
}
slow.Next = slow.Next.Next
return head
}
