Java删除链表的倒数第N个结点

19. 删除链表的倒数第N个结点

解题思路1:
链表的长度len
删除链表的倒数第N个结点就是删除链表的第len - N个结点

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode tmp = new ListNode(0, head);
        int len = 0;
        while(tmp.next != null){
            len++;
            tmp = tmp.next;
        }
        ListNode node = new ListNode(0, head);
        int count = 1;
        while(true){
            if(n == len){
                node.next = head.next;
                break;
            }else if(count == len - n){
                head.next = head.next.next;
                break;
            }
            head = head.next;
            count++;
        }
        return node.next;
    }
}

解题思路2:
利用栈先进后出

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0, head);
        Deque<ListNode> stack = new LinkedList<>();
        ListNode cur = dummy;
        while(cur != null){
            stack.push(cur);
            cur = cur.next;
        }
        for(int i = 0; i < n; i++){
            stack.pop();
        }
        ListNode pre = stack.peek();
        pre.next = pre.next.next;
        return dummy.next;
    }
}

解题思路3:
快慢指针
让快指针先走n步

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode fast = head, slow = head;
        for(int i = 0; i < n; i++){
            fast = fast.next;
        }
        //如果fast == null, 说明fast走n步走到了最后
        //说明要删除的是头结点
        if(fast == null){
            return head.next;
        }
        while(fast.next != null){
            fast = fast.next;
            slow = slow.next;
        }
        slow.next = slow.next.next;
        return head;
    }
}