Remove Nth Node From End of List

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.

解题思路：采用双指针，第一个指针p先走n-1步，在双指针同步，采用tmp指针保存要删除的前一个节点。

/*
* 边界输入：
* {1}, 1
* {1,2}, 1
* {1,2}, 2
*/

/**
* 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(head == NULL)
{
return NULL;
}
ListNode *p = head;
ListNode *s = head;
ListNode *tmp = NULL;
for(int i = 0; i < n - 1; i++)
{
if(p->next)
{
p = p->next;
}
}
while(p->next)
{
tmp = s;
p = p->next;
s = s->next;
}
if(tmp == NULL)
{
head = s->next;
delete s;
}
else
{
tmp->next = s->next;
delete s;
}
return head;
}
};