[LeetCode OJ]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.

开始时候自己想是要从后往前找，想得到个长度，不过后来又一想，链表怎么size。。。

还是双指针靠谱，一前一后，后面的比前面的大n-1。

/**
* 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 *temp = NULL;
ListNode *first = head;
ListNode *second = head;

for(int i = 1; i < n; i++)
first = first->next;

while(first->next) {
temp = second;
first = first->next;
second = second->next;
}

if(temp == NULL) {
head = second->next;
delete second;
}

else {
temp->next = second->next;
delete second;
}
return head;
}
};