【LeetCode】 19. Remove Nth Node From End of List 解题小结

题目：

Given a linked list, remove the nth node from the end of list and return its head.

For example,

  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.

首先容易想到的是先遍历一次链表，计算链表长度L，然后再次遍历到L-n的位置删除结点。更好的办法是应用递归，从后往前n递减，到-1时就到了要被删除结点的前一个结点。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

class Solution {

public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* node = new ListNode(0);
        node->next = head;
        removeNode(node, n);
        return node->next;
    }
    
    void removeNode(ListNode* head, int& n){
        if (!head) return;
        removeNode(head->next, n);
        n--;
        if (n == -1)
            head->next = head->next->next;
        return;
    }
};