leetcode:单链表之Remove Nth Node From End of List

leetcode:单链表之Remove Nth Node From End of List

题目:

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.

即:删除链表的倒数第n个元素

c++实现：

#include <iostream>

using namespace std;
struct ListNode
{
int val;
ListNode *next;
ListNode (int x):val(x),next(NULL){ }
};
ListNode *createListNode(int arr[],int  n)
{
ListNode *r;
ListNode *p;
ListNode * L=(ListNode*)malloc(sizeof(ListNode));

r=L;

for(int i=0;i<n;i++)
{
p=(ListNode*)malloc(sizeof(ListNode));
p->val=arr[i];
r->next=p;
r=p;
}
r->next=NULL;

return L->next;
}
ListNode *removeNthFromEnd(ListNode *head, int n)
{
if(head == NULL || head->next == NULL)
return NULL;

ListNode * first = head;
ListNode * second = head;

for(int i = 0;i < n;i++)  // first 先走n 步
first = first->next;
if(first == NULL)
return head->next;
while(first->next != NULL)  // 一起走
{
first = first->next;
second = second->next;
}
second->next = second->next->next;

return head;
}
int main()
{
int a[]={1,2,3,4,5};

ListNode *input;
ListNode *out;
int n=2;

input= createListNode(a,5);
out=removeNthFromEnd(input,2);
while(out != NULL)
{
cout<<out->val;
out = out->next;
}
cout<<endl;
return 0;
}

输出结果：