leetcode19

class Solution {
public:
    int N = ;
    int LEN = ;

    ListNode* removeNthFromEnd(ListNode* head, int n) {
        int i = ;
        ListNode* node = head;
        N++;
        LEN = max(LEN, N);
        if (node->next != NULL)
        {
            ListNode* Next = removeNthFromEnd(node->next, n);
            N--;
            if (n == LEN - N)
            {
                node->next = Next->next;
            }
        }
        if (n == LEN)
        {
            return head->next;
        }
        return node;
    }
};

补充一个python的实现：

 class Solution:
     N = 0
     LEN = 0
     def removeNthFromEnd(self, head: 'ListNode', n: 'int') -> 'ListNode':
         self.N += 1
         self.LEN = max(self.N,self.LEN)

         CN = self.N - 1
         node = head
         if node.next != None:
             nextnode = self.removeNthFromEnd(node.next,n)
         index = self.LEN - n - 1
         if CN == index:
             node.next = nextnode.next
         if n == self.LEN:
             return head.next
         return node

另一种快慢指针的实现不需要递归：

 public ListNode removeNthFromEnd(ListNode head, int n) {
     ListNode dummy = new ListNode();
     dummy.next = head;
     ListNode first = dummy;
     ListNode second = dummy;
     // Advances first pointer so that the gap between first and second is n nodes apart
     for (int i = ; i <= n + ; i++) {
         first = first.next;
     }
     // Move first to the end, maintaining the gap
     while (first != null) {
         first = first.next;
         second = second.next;
     }
     second.next = second.next.next;
     return dummy.next;
 }

python版本：

 class Solution:
     def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
         dm = ListNode(0)
         dm.next = head
         slow,fast = dm,dm
         for i in range(n):
             fast = fast.next
         while fast != None and fast.next != None:
             slow = slow.next
             fast = fast.next
         if slow.next != None:
             slow.next = slow.next.next
             return dm.next
         return None