README_EN.md

02.02. Kth Node From End of List

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Description

Implement an algorithm to find the kth to last element of a singly linked list. Return the value of the element.

Note: This problem is slightly different from the original one in the book.

Example:

Input:  1->2->3->4->5 和 k = 2

Output:  4

Note:

k is always valid.

Solutions

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:
        p = q = head
        for _ in range(k):
            q = q.next
        while q:
            p, q = p.next, q.next
        return p.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 p = head, q = head;
        while (k-- > 0) {
            q = q.next;
        }
        while (q != null) {
            q = q.next;
            p = p.next;
        }
        return p.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;
};

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