面试题 22. 链表中倒数第 k 个节点

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。

Python3JavaC++GoRustJavaScriptC#

# 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

/**
 * 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;
    }
}

/**
 * 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;
    }
};

/**
 * 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
}

// 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()
    }
}

/**
 * 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;
};

/**
 * 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;
    }
}

面试题 22. 链表中倒数第 k 个节点

题目描述

解法

方法一：快慢指针

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