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个结点。

下面是我第一次写的蠢代码 复杂度 n+2k

/*
public class ListNode {
    int val;
    ListNode next = null;


    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode FindKthToTail(ListNode head,int k) {
        if (head == null) {
return head;
}
        if (k <1) {
            return null;
        }
ListNode qian = null;
int len = 1;
while (head.next != null) {
ListNode ing = head.next;
head.next = qian;
qian = head;
head = ing;
len ++;
}
if (len < k) {
return null;
}
head.next = qian;
qian = null;
while (-- k > 0) {
ListNode ing = head.next;
head.next = qian;
qian = head;
head = ing;
}
head.next = qian;
return head;
    }
}

这是我参考别人的写的代码 时间复杂度n 诶跟别人比差的很远 我还是太菜

class ListNode {
    int val;
    ListNode next = null;


    ListNode(int val) {
        this.val = val;
    }
}
public class Solution {
    public ListNode FindKthToTail(ListNode head,int k) {
        if (k < 1 || head == null) {
return null;
}
        ListNode a = head;
        ListNode b = head;
        while (--k > 0) {
        b = b.next;
        if(b == null) return null;
        }
        while (b.next != null) {
        a = a.next;
        b = b.next;
        }
        return a;
    }
}