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面试题 02.02. 返回倒数第 k 个节点

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 \(O(1)\)。

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# 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

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/**
 * 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;
    }
}

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/**
 * 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;
    }
};

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/**
 * 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
}

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/**
 * 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;
}

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// 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
    }
}

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/**
 * 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;
};

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     var val: Int
 *     var next: ListNode?
 *     init(_ x: Int, _ next: ListNode? = nil) {
 *         self.val = x
 *         self.next = next
 *     }
 * }
 */

class Solution {
    func kthToLast(_ head: ListNode?, _ k: Int) -> Int {
        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?.val ?? 0
    }
}

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