Lintcode Delete the nth node in a linked list

Given a linked list, delete the nth node in the list and return the head node of the linked list.

Sample Example

Give the list 1->2->3->4->5->null and n = 2.

After you delete the second-to-last node, the list becomes 1->2->3->5->null.

Analysis: Considering the robustness of each case, it is necessary to consider the tail node, especially when the tail node is deleted.

/** * Definition of ListNode * class ListNode {* Public: * int val; * ListNode *next; * ListNode (int val) {* This->val = val; * This->next = NULL;     *} *} */class Solution {public:/** * @param head:the first node of the linked list.     * @param n:an Integer.     * @return: The head of linked list. */ListNode *removenthfromend (listnode *head, int n) {//write your code here if (head==null) Retu        RN 0;        if (head->next==null&&n==1) return 0;        ListNode *p1=head;        ListNode *p2=null;        ListNode *pre=null;            for (int i=0;i<n-1;i++) {if (p1->next!=null) {p1=p1->next;        } else return NULL;        } P2=head;            while (p1->next!=null) {p1=p1->next;            PRE=P2;        p2=p2->next; } if (p2->next==null) {Pre->neXt=null;            Delete P2;        P2=null;            } else {ListNode *p3=p2->next;            p2->val=p3->val;            p2->next=p3->next;            Delete P3;        P3=null;    } return head; }};

　　

Lintcode Delete the nth node in a linked list