leetcode: 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.

思路

把所有的指针全部存下来，想不到其他的只遍历一次的方法了。自己中间马虎，提交了好几次才成功

。一定警告自己注意边界条件。。

/**
* 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) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
vector<ListNode *> pt;
ListNode * cur = head;
while(cur != NULL)
{
pt.push_back(cur);
cur = cur->next;
}

if(pt.size() == 0) return head;
if(n <= 0 || n > pt.size()) return head;
if(pt.size() == 1)
{
delete head;
return NULL;
}
if(pt.size() == n)
{
cur = head->next;
head->next = NULL;
delete head;
return cur;
}
if(n == 1)
{
pt[pt.size()-2]->next = NULL;
delete pt[pt.size() - 1];
return head;
}
ListNode * pr = pt[pt.size() - n - 1];
cur = pr->next;
pr->next = cur->next;
cur->next = NULL;
delete cur;
return head;
}
};