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中等
链表
双指针

19. 删除链表的倒数第 N 个结点

English Version

题目描述

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。

 

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：你能尝试使用一趟扫描实现吗？

解法

方法一：快慢指针

我们定义两个指针 fast 和 slow，初始时都指向链表的虚拟头结点 dummy。

接着 fast 指针先向前移动 $n$ 步，然后 fast 和 slow 指针同时向前移动，直到 fast 指针到达链表的末尾。此时 slow.next 指针指向的结点就是倒数第 n 个结点的前驱结点，将其删除即可。

时间复杂度 $O(n)$，其中 $n$ 为链表的长度。空间复杂度 $O(1)$。

Python3

# 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: Optional[ListNode], n: int) -> Optional[ListNode]:
        dummy = ListNode(next=head)
        fast = slow = dummy
        for _ in range(n):
            fast = fast.next
        while fast.next:
            slow, fast = slow.next, fast.next
        slow.next = slow.next.next
        return dummy.next

Java

/**
 * 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;
    }
}

C++

/**
 * 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;
    }
};

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func removeNthFromEnd(head *ListNode, n int) *ListNode {
	dummy := &ListNode{0, head}
	fast, slow := dummy, dummy
	for ; n > 0; n-- {
		fast = fast.Next
	}
	for fast.Next != nil {
		slow, fast = slow.Next, fast.Next
	}
	slow.Next = slow.Next.Next
	return dummy.Next
}

TypeScript

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
    const dummy = new ListNode(0, head);
    let fast = dummy;
    let slow = dummy;
    while (n--) {
        fast = fast.next;
    }
    while (fast.next) {
        slow = slow.next;
        fast = fast.next;
    }
    slow.next = slow.next.next;
    return dummy.next;
}

Rust

// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn remove_nth_from_end(head: Option<Box<ListNode>>, n: i32) -> Option<Box<ListNode>> {
        let mut dummy = Some(Box::new(ListNode { val: 0, next: head }));
        let mut slow = &mut dummy;
        let mut fast = &slow.clone();
        for _ in 0..=n {
            fast = &fast.as_ref().unwrap().next;
        }
        while fast.is_some() {
            fast = &fast.as_ref().unwrap().next;
            slow = &mut slow.as_mut().unwrap().next;
        }
        slow.as_mut().unwrap().next = slow.as_mut().unwrap().next.as_mut().unwrap().next.take();
        dummy.unwrap().next
    }
}

JavaScript

/**
 * 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;
};

Swift

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func removeNthFromEnd(_ head: ListNode?, _ n: Int) -> ListNode? {
        let dummy = ListNode(0)
        dummy.next = head
        var fast: ListNode? = dummy
        var slow: ListNode? = dummy
        for _ in 0..<n {
            fast = fast?.next
        }
        while fast?.next != nil {
            fast = fast?.next
            slow = slow?.next
        }
        slow?.next = slow?.next?.next
        return dummy.next
    }
}
```

#### Ruby

```rb
# 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
```

#### PHP

```php
/**
 * Definition for a singly-linked list.
 * class ListNode {
 *     public $val = 0;
 *     public $next = null;
 *     function __construct($val = 0, $next = null) {
 *         $this->val = $val;
 *         $this->next = $next;
 *     }
 * }
 */
class Solution {
    /**
     * @param ListNode $head
     * @param Integer $n
     * @return ListNode
     */
    function removeNthFromEnd($head, $n) {
        $dummy = new ListNode(0, $head);
        $fast = $slow = $dummy;
        for ($i = 0; $i < $n; $i++) {
            $fast = $fast->next;
        }
        while ($fast->next !== null) {
            $fast = $fast->next;
            $slow = $slow->next;
        }
        $slow->next = $slow->next->next;
        return $dummy->next;
    }
}
```

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