README.md

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

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。

Python3

# 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

Java

/**
 * 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;
    }
}

JavaScript

/**
 * 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;
};

C++

/**
 * 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;
    }
};

Go

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
}

...