LeetCode 19. Remove Nth Node From End of List--删除链表的倒数第n个结点

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

class ListNode {
int val;
ListNode next;

ListNode(int x) {
val = x;
}
}

public class Main {
//方法一：遍历两遍
public ListNode removeNthFromEnd1(ListNode head, int n) {
int size = 0;
ListNode curNode = head;
while (curNode != null) {//获取链表长度
size++;
curNode = curNode.next;
}

if (size == 1 && n == 1) {
return null;
}

if (n == size) {
return head.next;
}

if (n > size) {
return null;
}
ListNode preNode = null;
preNode = head;
for (int i = 0; i < size - n - 1; i++) {
preNode = preNode.next;
}
preNode.next = preNode.next.next;
return head;
}

//方法二：遍历一遍,两个指针，一前一后，
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode start = new ListNode(0);//需要加一个头结点，比如1,2删除倒数第2个结点，需要把1删掉，此时1之前没有结点很难处理
start.next = head;
ListNode fastPoint = start;
ListNode slowPoint = start;

for (int i = 0; i < n + 1; i++) {//下标的关系：slowPoint+n+1 = fastPoint，比如1+2+1 = 4
fastPoint = fastPoint.next;
}
while (fastPoint != null) {//当fastPoint指向9时，还能继续下次next，
fastPoint = fastPoint.next;//此时fastPoint.next = null，slowPoint.next=7，然后才跳出循环
slowPoint = slowPoint.next;
}
slowPoint.next = slowPoint.next.next;//7->8->9变为7->9
//        return head;//不能再是返回head
return start.next;//slowPoint与start指向的是同一个地址，return slowPoint.next;也正确
}

public static void main(String[] args) {
ListNode node1 = new ListNode(1);
node1.next = new ListNode(2);
//        node1.next.next = new ListNode(3);
//        node1.next.next.next = new ListNode(4);
//        node1.next.next.next.next = new ListNode(5);
//        node1.next.next.next.next.next = new ListNode(6);
//        node1.next.next.next.next.next.next = new ListNode(7);
//        node1.next.next.next.next.next.next.next = new ListNode(8);
//        node1.next.next.next.next.next.next.next.next = new ListNode(9);
//        new Main().removeNthFromEnd(node1, 2);

System.out.println(new Main().removeNthFromEnd(node1, 2).val);//2
}
}