题目描述
输入一个链表,输出该链表中倒数第k个节点。为了符合大多数人的习惯,本题从1开始计数,即链表的尾节点是倒数第1个节点。
例如,一个链表有 6
个节点,从头节点开始,它们的值依次是 1、2、3、4、5、6
。这个链表的倒数第 3
个节点是值为 4
的节点。
示例:
给定一个链表: 1->2->3->4->5 , 和 k = 2 .
返回链表 4->5 .
解法
方法一:快慢指针
我们可以定义快慢指针 fast
和 slow
,初始时均指向 head
。
然后快指针 fast
先向前走 $k$ 步,然后快慢指针同时向前走,直到快指针走到链表尾部,此时慢指针指向的节点就是倒数第 $k$ 个节点。
时间复杂度 $O(n)$,空间复杂度 $O(1)$。其中 $n$ 为链表长度。
Python3 Java C++ Go Rust JavaScript C# Swift
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15 # Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution :
def getKthFromEnd ( self , head : ListNode , k : int ) -> ListNode :
slow = fast = head
for _ in range ( k ):
fast = fast . next
while fast :
slow , fast = slow . next , fast . next
return slow
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21 /**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode getKthFromEnd ( ListNode head , int k ) {
ListNode slow = head , fast = head ;
while ( k -- > 0 ) {
fast = fast . next ;
}
while ( fast != null ) {
slow = slow . next ;
fast = fast . next ;
}
return slow ;
}
}
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22 /**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public :
ListNode * getKthFromEnd ( ListNode * head , int k ) {
ListNode * slow = head , * fast = head ;
while ( k -- ) {
fast = fast -> next ;
}
while ( fast ) {
slow = slow -> next ;
fast = fast -> next ;
}
return slow ;
}
};
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17 /**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func getKthFromEnd ( head * ListNode , k int ) * ListNode {
slow , fast := head , head
for ; k > 0 ; k -- {
fast = fast . Next
}
for fast != nil {
slow , fast = slow . Next , fast . Next
}
return slow
}
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30 // 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 get_kth_from_end ( head : Option < Box < ListNode >> , k : i32 ) -> Option < Box < ListNode >> {
let mut fast = & head ;
for _ in 0 .. k {
fast = & fast . as_ref (). unwrap (). next ;
}
let mut slow = & head ;
while let ( Some ( nf ), Some ( ns )) = ( fast , slow ) {
fast = & nf . next ;
slow = & ns . next ;
}
slow . to_owned ()
}
}
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24 /**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @param {number} k
* @return {ListNode}
*/
var getKthFromEnd = function ( head , k ) {
let fast = head ;
while ( k -- ) {
fast = fast . next ;
}
let slow = head ;
while ( fast ) {
slow = slow . next ;
fast = fast . next ;
}
return slow ;
};
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21 /**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode GetKthFromEnd ( ListNode head , int k ) {
ListNode fast = head , slow = head ;
while ( k -- > 0 ) {
fast = fast . next ;
}
while ( fast != null ) {
slow = slow . next ;
fast = fast . next ;
}
return slow ;
}
}
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29 /* public class ListNode {
* var val: Int
* var next: ListNode?
* init(_ val: Int) {
* self.val = val
* self.next = nil
* }
* }
*/
class Solution {
func getKthFromEnd ( _ head : ListNode ?, _ k : Int ) -> ListNode ? {
var slow = head
var fast = head
var k = k
while k > 0 {
fast = fast ?. next
k -= 1
}
while fast != nil {
slow = slow ?. next
fast = fast ?. next
}
return slow
}
}