leetcode19

class Solution {
public:
    int N = 0;
    int LEN = 0;

    ListNode* removeNthFromEnd(ListNode* head, int n) {
        int i = 0;
        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(0);
    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 = 1; i <= n + 1; 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