剑指 Offer 22. 链表中倒数第k个节点🌟🌟🌟🌟🌟中等

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问题描述

原文链接：剑指 Offer 22. 链表中倒数第k个节点

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.

代码实现

Java

/**
 * 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) {
        if(head == null){
            return head;
        }

        ListNode slow = head;
        ListNode fast = head;

        for(int i = 0; i < k - 1; i++){
            fast = fast.next;
        }

        while(fast.next != null){
            fast = fast.next;
            slow = slow.next;
        }

        return slow;
    }
}

Python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def getKthFromEnd(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if head is None:
            return head

        slow = head
        fast = head

        for i in range(k - 1):
            fast = fast.next

        while fast.next is not None:
            fast = fast.next
            slow = slow.next

        return slow

C++

/**
 * 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) {
        if(head == nullptr){
            return head;
        }

        ListNode* slow = head;
        ListNode* fast = head;

        for(int i = 0; i < k - 1; i++){
            fast = fast->next;
        }

        while(fast->next != nullptr){
            fast = fast->next;
            slow = slow->next;
        }

        return slow;
    }
};

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func getKthFromEnd(head *ListNode, k int) *ListNode {
    if head == nil {
        return head
    }

    slow := head
    fast := head

    for i := 0; i < k-1; i++ {
        fast = fast.Next
    }

    for fast.Next != nil {
        fast = fast.Next
        slow = slow.Next
    }

    return slow
}

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