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;}