【Leetcode长征系列】Remove Nth Node From End of List

原题：

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.

思路：用两个指针，让它们相隔n，当后一个指针指到最尾端时，前一个刚好指到我们需要删除的点。不过要一个额外的指针记录删除点的前一个节点。还需要考虑到删除头结点和末尾节点的情况。

/**
* 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 *mark = NULL;
ListNode *tmp = head;
ListNode *del = head;

if (!head||!head->next) return NULL;
for(int i=0; i<n-1; i++)
tmp = tmp->next;

while(tmp->next){
mark = del;
del = del->next;
tmp = tmp->next;
}
if (mark==NULL){
head=del->next;
delete del;
}
else{
mark->next = del->next;
delete del;
}
return head;
}
};