019 Remove Nth Node From End of List [Leetcode]

題目內容：

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.

思路：
使用雙指針，讓前置指針先走n步，然後前後兩個指針一起向前直到前置指針到達鏈表尾部，刪除後指針指向的節點。
同樣使用了附加頭結點來簡化運算。（如刪除的節點是第一個節點的時候）

代碼：

/**
 * 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) {
        if (n == 0 || head == NULL)
            return head;

        ListNode *pre_head = new ListNode(0);
        pre_head->next = head;

        // let first point take n steps, and let two points go to the end
        ListNode *front(pre_head), *end(pre_head), *pre(pre_head);
        for (int i = 0; i < n; ++i) {
            front = front->next;
        }

        while(front != NULL) {
            pre = end;
            front = front->next;
            end = end->next;
        }

        // delete node
        pre->next = end->next;
        end->next = NULL;
        delete end;

        // delete pre-head point
        head = pre_head->next;
        pre_head->next = NULL;
        delete pre_head;
        return head;
    }
};