LeetCode 19. 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.

题目要求我们要尽量扫描一遍，那我们就要设置两个指针了，第二个指针先前行n+1，然后第一个指针再前行，当第二个指针到达结尾时，第一个指针就是

要删除节点的前序，同时也要考虑恰好头结点就是要删除节点的特殊情况

/**
* 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 *pre = head, *current = head;
int count = 0;
while (current)
{
if (!(count <= n))
pre = pre->next;
++count;
current = current->next;
}
if (pre == head&&count==n)
{
head = head->next;
free(pre);
}
else
{
current = pre->next;
pre->next = pre->next->next;
free(current);
}
return head;
}
};