剑指offer 链表中倒数第k个结点

题目：

输入一个链表，输出该链表中倒数第k个结点

思路：主要考虑   1. head为空     2.链表长度小于k     3.k=0 无符号减1会变成4294967295(Java有没有?)

public class Solution {
    
    //两个指针，一前一后
    public ListNode FindKthToTail(ListNode head,int k) {
        if(head == null || k <= 0){
            return null;
        }

        ListNode pre = head;  //先走k-1个指针
        ListNode last = head;

        for(int i = 1; i < k; i++){
            if(pre.next != null){
                pre = pre.next;
            }else {
                return null;
            }
        }

        while(pre.next != null){
            pre = pre.next;
            last = last.next;
        }
        return last;
    }

    //栈，但是空间复杂度，当链表特别长，容易出现Stack Overflow问题
    public ListNode FindKthToTail(ListNode head,int k) {
        if(head == null || k == 0){
            return null;
        }

        Stack<ListNode> stack = new Stack<ListNode>();
        int count = 0;
        while(head != null){
            stack.push(head);
            head = head.next;
            count ++;
        }
        if(count < k){
            return null;
        }

        ListNode node  = null;
        for(int i = 0; i < k; i++){
            node = stack.pop();
        }
        return node;
    }


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

        ListNode(int val) {
            this.val = val;
        }
    }

}