Question

面试题 02.02. 返回倒数第 k 个节点

English Version

题目描述

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。

解法

方法一：快慢指针

我们定义两个指针 slow 和 fast，初始时都指向链表头节点 head。然后 fast 指针先向前移动 $k$ 步，然后 slow 和 fast 指针同时向前移动，直到 fast 指针指向链表末尾。此时 slow 指针指向的节点就是倒数第 $k$ 个节点。

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

# 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 = slow.next
      fast = 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.
 * 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--) {
      fast = fast->next;
    }
    while (fast) {
      slow = slow->next;
      fast = fast->next;
    }
    return slow->val;
  }
};

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

/**
 * 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 kthToLast(head: ListNode | null, k: number): number {
  let [slow, fast] = [head, head];
  while (k--) {
    fast = fast.next;
  }
  while (fast !== null) {
    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
  }
}

/**
 * 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 [slow, fast] = [head, head];
  while (k--) {
    fast = fast.next;
  }
  while (fast !== null) {
    slow = slow.next;
    fast = fast.next;
  }
  return slow.val;
};