LeetCode 19：删除链表的倒数第N个节点 Remove Nth Node From End of List

给定一个链表: 1->2->3->4->5, 和 n = 2.

当删除了倒数第二个节点后，链表变为 1->2->3->5.

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode curA = head;
        ListNode curB = head;
        for (int i = 0; i < n; i++) curA = curA.next;
        if (curA == null) {//如果走了n步之后该节点指向空节点，则该链表只有一个节点
            head = head.next;
            return head;
        }
        while (curA.next != null) {//当第一个指针的下一个节点为空时，该指针指向最后一个节点，而指针curB 走了L-n-1步，即指向该删除节点的前一个节点
            curA = curA.next;
            curB = curB.next;
        }
        curB.next = curB.next.next;//将本来指向应当删除节点地址指向应当删除节点的下一个节点的地址
        return head;
    }
}

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        curA,curB=head,head
        for i in range(n):
            curA=curA.next
        if not curA:
            head=head.next
            return head
        while(curA.next):
            curA=curA.next
            curB=curB.next
        curB.next=curB.next.next