[leetcode]40 Remove Nth Node From End of List

题目链接：https://leetcode.com/problems/remove-nth-node-from-end-of-list/

Runtimes：11ms

1、问题

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.

2、分析

先扫描一遍，知道总长度length，在往后移动length-n个节点，删除，over。

3、小结

一般链表的节点删除需要移动到删除节点的前一个节点，因此如果删除节点是头结点，处理的方法和其他的不同，这是需要注意的。

4、实现

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

5、反思

效果不错。