Remove Nth Node From End of List —— Leetcode

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.

题目很容易理解，即删除倒数第n个结点，注意以下两个地方：

（1）要求只遍历一次，一个指针不行的情况下，要想到用两个指针；

（2）链表删除问题边界情况都有头结点的问题，所以要注意这里的边界情况（注释地方为第一次提交写错的地方）

C代码：

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode *removeNthFromEnd(struct ListNode *head, int n) {
struct ListNode *i=head, *j=head, *pi=head;
for(int k=1; k<n; k++)
j = j->next;
while(j->next != NULL)
{
pi = i;
i = i->next;
j = j->next;
}
if(i == head) // not if(pi == head)
{
head = head->next;
pi->next = NULL;
}
else
{
pi->next = i->next;
i->next = NULL;
}

return head;
}