leetcode19---Remove Nth Node From End of List

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.

#include <iostream>
#include<stdlib.h>
using namespace std;

struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
if(head == NULL || head->next == NULL) return NULL;
int length=0;//一共有多少个元素
ListNode* p1=head;
while(p1 != NULL)
{
p1=p1->next;
length++;
}
//cout<<length<<endl;
int k=length-n+1;//得到正序的位置
//cout<<k<<endl;
if(k==1) return head->next;//(1)如果删除的是头结点直接返回head->next
ListNode* p2=head;
while(k-- != 2)//(2)如果删除的不是头结点，找到要删除节点的前一个节点
{
p2=p2->next;
}
p2->next=p2->next->next;//删除
return head;
}
ListNode* CreateList(int a[], int n)
{//头结点包含数据的情况
ListNode *head, *p, *q;

head = (ListNode*)malloc(sizeof(ListNode));
head->next = NULL;

p = head;
p->val = a[0];
//p->next = NULL;
for(int i=1; i<n; i++)
{
q = (ListNode*)malloc(sizeof(ListNode));
q->val = a[i];//数组元素给链表赋值！！！
p->next = q;
p = q;
}
p->next = NULL;//将最后一个结点的指针域清空
return head;//返回这个链表的首地址
}
};
int main()
{
int a[]={1,2,3,4,5};
//int a2[]={1};
Solution s;
ListNode *head=s.CreateList(a, 5);
head=s.removeNthFromEnd(head, 2);
//ListNode *p=head->next;
ListNode *p=head;
while(p != NULL)
{
cout<<p->val<<endl;
p=p->next;
}
return 0;
}

1
2
3
5

Process returned 0 (0x0)   execution time : 0.027 s
Press any key to continue.