移除链表的第n个节点

利用快慢指针来实现，快指针先执行头节点的第n个节点，慢指针指向头节点

然后快慢指针一起移动，当快指针指向尾节点时，慢指针的下一个节点就是要移除的节点

代码实现：

public class Client {
    public static void main(String[] args) {
        ListNode head = new ListNode(0);
        ListNode curNode = head;
        for (int i = 1; i < 5; i++) {
            ListNode nextNode = new ListNode(i);
            curNode.setNext(nextNode);
            curNode = nextNode;
        }

        removeNthFromEnd(head,3);
    }

    /**
     * 移除单链表的第n个节点
     */
    static void removeNthFromEnd(ListNode head, int n) {
        ListNode fast = head;
        ListNode slow = head;
        ListNode tempNode = head;
        for (int i = 0; i < n; i++) {
            fast = tempNode.getNext();
            tempNode = tempNode.getNext();
        }

        while (fast.getNext() != null){
            fast=fast.getNext();
            slow = slow.getNext();
        }
        slow.setNext(slow.getNext().getNext());
    }
}