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**.
Solution:
/**
* 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) {
if(NULL == head) return head;
int size = 0;
for(ListNode* p = head;p!=NULL;p=p->next) size++;
if(n == size) return head->next;
ListNode *p = head;
for(int i=0;i<size - n - 1;i++) p = p->next;
p->next = p->next->next;
return head;
}
};
