leetcode:單鏈表之Remove Nth Node From End of List

leetcode:單鏈表之Remove Nth Node From End of List

題目:

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:刪除鏈表的倒數第n個元素

c++實現：

#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;
}

測試結果：