leetcode 19: Remove Nth Node From End of List

问题描述：

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：

题目看来挺简单，主要就是设置一个前行节点一个后行节点，同步控制两节点距离一直到链表末端。但是实践完成代码的思路上还是有一点点技巧和注意的。
这里我用了一个链表附加头节点，这样在删除头结点问题上可以很方便解决；然后注意要得到的节点应该是所要求删除节点的前一个节点；还有就是最后返回链表时，要返回更新后的head节点，因为head可能有改变。

代码：

/**
 * 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* prehead = new ListNode(-1);
        prehead->next = head;
        ListNode* bias = prehead;
        ListNode* targetpre = prehead;
        //bias move forward n steps
        for(int i = 0; i < n; i++)  
        {
            if(bias == NULL) return head;
            bias = bias->next;   
        }
        //bias point to the end, at the same time, target point to the previous node of the nth from the end.
        while(bias->next != NULL)
        {
            bias = bias->next;
            targetpre = targetpre->next;
        }

        ListNode* temp = targetpre->next;
        targetpre->next = temp->next;
        head = prehead->next;   //update the new head
        delete temp;
        delete prehead;
        return head;
    }
};