[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.

/**
* 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* head0 = new ListNode(0);//需要注意的是，这题的head是第一个节点，不是通常所说的指向第一个节点的head，所以要建个头节点以便于操作
head0 -> next = head;
ListNode* p1 = head0;
ListNode* p2 = head0;
for(int i = 0;i < n;i ++)
p2 = p2 -> next;
while(p2 -> next){
p1 = p1 -> next;
p2 = p2 -> next;
}

p1 -> next = p1 ->next -> next;

return head0 -> next;
}
};