19. Remove Nth Node From End of List

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.

Note:

Given n will always be valid.
Try to do this in one pass.

代码如下：

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        vector<int> value;
        ListNode* p = head;
        while (p != NULL) {
          value.push_back(p -> val);
          p = p -> next;
        }
        if (value.size() == 0)
          return head;
        if (value.size() == n) {
          ListNode* tmp = head;
          head = tmp -> next;
          delete tmp;
          return head;
        }
        ListNode* t = head;
        for (int i = 0; i < value.size() - n - 1; i++) 
          t = t -> next;
        ListNode* m = t -> next;
        t -> next = m -> next;
        delete m;
        return head;
    }
};

解题思路：

　　考虑三种情况：当链表为空时、当删除中间和末尾元素时、当删除第一个元素时（注意头指针ｈｅａｄ的变化）。