给定一个链表: 1->2->3->4->5, 和 k = 2.
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返回链表 4->5.

//方法1借助LinkedList
class Solution {
    //遍历将每一个都添加到头部
    public ListNode getKthFromEnd(ListNode head, int k) {
        LinkedList<ListNode> list = new LinkedList<>();
        while(head!=null){
            list.offerFirst(head);
            head = head.next;
        }
        return (k > 0 && list.size() >= k)?list.get(k-1):null;
    }
}

class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        ListNode slow = head;
        ListNode fast = head;
        //快指针先走k步 
        //i从0开始所以小于k
        for(int i = 0;i < k;i++){
            fast = fast.next;
        }
        while(fast!=null){
            slow = slow.next;
            fast = fast.next;
        }
        return slow;
    }
}