Remove Nth Node From End of List

id19. Remove Nth Node From End of List Add to List QuestionEditorial Solution My Submissions
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Total Submissions: 480607
Difficulty: Easy
Contributors: Admin
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.

主要使用了双指针，比较重要得一点是建立头节点，因为是一个独立结点所以需要分配空间

/**
 * 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* result=NULL;
    if(head==NULL||head->next==NULL)
    return result;
   ListNode* Lhead= new ListNode(sizeof(ListNode));
//  Lhead->val=99;
   Lhead->next=head;
  // cout<<Lhead->next->next->val;
        ListNode *p=Lhead;
        for(int i=0;i<=n;i++)
        {
        p=p->next;

        }
        ListNode *q=Lhead;
        while(p)
        {p=p->next;
         q=q->next;
        }
        ListNode* h=q->next->next;

    //  cout<<h->val<<q->val;
        q->next=h;
     //   cout<<q->val;
        result=Lhead->next;
        return result;
    }
};