实现一种算法,找出单向链表中倒数第 k 个节点。返回该节点的值。
注意:本题相对原题稍作改动
示例:
输入: 1->2->3->4->5 和 k = 2 输出: 4
说明:
给定的 k 保证是有效的。
定义 p
、q
指针指向 head
。
p
先向前走 k
步,接着 p
、q
同时向前走,当 p
指向 null
时,q
指向的节点即为链表的倒数第 k
个节点。
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def kthToLast(self, head: ListNode, k: int) -> int:
slow = fast = head
for _ in range(k):
fast = fast.next
while fast:
slow, fast = slow.next, fast.next
return slow.val
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public int kthToLast(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.val;
}
}
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @param {number} k
* @return {number}
*/
var kthToLast = function (head, k) {
let fast = head,
slow = head;
for (let i = 0; i < k; i++) {
fast = fast.next;
}
while (fast != null) {
fast = fast.next;
slow = slow.next;
}
return slow.val;
};
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
int kthToLast(ListNode* head, int k) {
ListNode* fast = head;
ListNode* slow = head;
while (k-- > 0) {
fast = fast->next;
}
while (fast) {
slow = slow->next;
fast = fast->next;
}
return slow->val;
}
};
func kthToLast(head *ListNode, k int) int {
slow, fast := head, head
for i := 0; i < k; i++ {
fast = fast.Next
}
for fast != nil {
slow = slow.Next
fast = fast.Next
}
return slow.Val
}
// 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 kth_to_last(head: Option<Box<ListNode>>, k: i32) -> i32 {
let mut fast = &head;
for _ in 0..k {
fast = &fast.as_ref().unwrap().next;
}
let mut slow = &head;
while let (Some(f), Some(s)) = (fast, slow) {
fast = &f.next;
slow = &s.next;
}
slow.as_ref().unwrap().val
}
}