[Leetcode] remove nth node from the end of list 删除链表倒数第n各节点

Given a linked list, remove the n th 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.

这题比较简单，使用快慢指针，找到倒数第n个结点的前驱即可，然后连接其前驱和后继即可，值得注意的是，两个while的条件之间的关系。代码如下：

/**
 * 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 *nList=new ListNode(-1);
        nList->next=head;
        ListNode *pre=nList;
        ListNode *fast=head;
        ListNode *slow=head;
        int num=0;

        while(num++<n)
        {
            fast=fast->next;
        }   
        while(fast->next)
        {
            pre=pre->next;
            slow=slow->next;
            fast=fast->next;
        } 
        pre->next=slow->next;

        return nList->next;
    }
};