题目描述
给定一个链表,删除链表的倒数第 n
个结点,并且返回链表的头结点。
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
进阶: 能尝试使用一趟扫描实现吗?
注意:本题与主站 19 题相同: https://leetcode.cn/problems/remove-nth-node-from-end-of-list/
解法
方法一
Python3 Java C++ Go JavaScript Ruby Swift
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16 # 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 :
dummy = ListNode ( next = head )
slow , fast = dummy , dummy
for _ in range ( n ):
fast = fast . next
while fast . next :
slow = slow . next
fast = fast . next
slow . next = slow . next . next
return dummy . next
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25 /**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode removeNthFromEnd ( ListNode head , int n ) {
ListNode dummy = new ListNode ( 0 , head );
ListNode fast = dummy , slow = dummy ;
while ( n -- > 0 ) {
fast = fast . next ;
}
while ( fast . next != null ) {
slow = slow . next ;
fast = fast . next ;
}
slow . next = slow . next . next ;
return dummy . next ;
}
}
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27 /**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public :
ListNode * removeNthFromEnd ( ListNode * head , int n ) {
ListNode * dummy = new ListNode ( 0 , head );
ListNode * fast = dummy ;
ListNode * slow = dummy ;
while ( n -- ) {
fast = fast -> next ;
}
while ( fast -> next ) {
slow = slow -> next ;
fast = fast -> next ;
}
slow -> next = slow -> next -> next ;
return dummy -> next ;
}
};
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22 /**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func removeNthFromEnd ( head * ListNode , n int ) * ListNode {
dummy := & ListNode { 0 , head }
fast := dummy
slow := dummy
for n > 0 {
fast = fast . Next
n -= 1
}
for fast . Next != nil {
slow = slow . Next
fast = fast . Next
}
slow . Next = slow . Next . Next
return dummy . Next
}
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26 /**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @param {number} n
* @return {ListNode}
*/
var removeNthFromEnd = function ( head , n ) {
const dummy = new ListNode ( 0 , head );
let fast = dummy ,
slow = dummy ;
while ( n -- ) {
fast = fast . next ;
}
while ( fast . next ) {
slow = slow . next ;
fast = fast . next ;
}
slow . next = slow . next . next ;
return dummy . next ;
};
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25 # Definition for singly-linked list.
# class ListNode
# attr_accessor :val, :next
# def initialize(val = 0, _next = nil)
# @val = val
# @next = _next
# end
# end
# @param {ListNode} head
# @param {Integer} n
# @return {ListNode}
def remove_nth_from_end ( head , n )
dummy = ListNode . new ( 0 , head )
fast = slow = dummy
while n > 0
fast = fast . next
n -= 1
end
while fast . next
slow = slow . next
fast = fast . next
end
slow . next = slow . next . next
return dummy . next
end
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30 /* class ListNode {
* var val: Int
* var next: ListNode?
* init() { self.val = 0; self.next = nil }
* init(_ val: Int) { self.val = val; self.next = nil }
* init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next }
* }
*/
class Solution {
func removeNthFromEnd ( _ head : ListNode ?, _ n : Int ) -> ListNode ? {
let dummy = ListNode ( 0 , head )
var fast : ListNode ? = dummy
var slow : ListNode ? = dummy
var n = n
while n > 0 {
fast = fast ?. next
n -= 1
}
while fast ?. next != nil {
slow = slow ?. next
fast = fast ?. next
}
slow ?. next = slow ?. next ?. next
return dummy . next
}
}