[Network] cattle off the list penultimate k nodes

First, the subject description:

Input a linked list, the linked list output reciprocal k-th node.

Second, the problem-solving ideas:

1, the length of chain required size.
2, the k-node is equal to the reciprocal of the number of positive (size-k) nodes.
3, over the interest list, go (size-k) step can be.

Third, Code Description:

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

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    public ListNode FindKthToTail(ListNode head,int k) {
       //判断k的合法性
        if(k<0||k>size){
            return null;
        }        
        int size=getSize(head);
        int steps=size-k;
        ListNode cur=head;
        for(int i=0;i<steps;i++){
            cur=cur.next;
        }
       return cur;
    }
    //获取链表长度
    private int getSize(ListNode head){
        int size=0;
        for(ListNode cur=head;cur!=null;cur=cur.next){
            size++;
        }
        return size;
    }
}