LeetCode算法题——19. 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.

算法思想：

先求解单链表的长度len，由此可知，所要删除的元素位于单链表的第len-n个位置，然后找到该位置，进行删除

C++实现：

#include 
#include 
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) {
    	ListNode *copyHead,*returnList,*p1,*p2;
    	copyHead=head;
    	returnList=NULL;
    	p1=NULL;
    	p2=NULL;
    	int len=0;
        while(copyHead!=NULL){
        	len++;
        	copyHead=copyHead->next;        	
        }
    	int i=0;
    	while(head!=NULL){
    		if(i!=(len-n)){
    			p2=(ListNode *)malloc(sizeof(ListNode));
    			p2->val=head->val;
    			p2->next=NULL;
    			if(returnList==NULL){
    				returnList=p1=p2;
    			}else{
    				p1->next=p2;
    				p1=p1->next;
    				p1->next=NULL;
    			}
    		}
			head=head->next;			
    		i++;
    	}
		return returnList;
    }
};
int main(){
	ListNode *head,*p1,*p2,*res;
	head=NULL;
	p1=NULL;
	p2=NULL;
	int a[5]={1,2,3,4,5};
	for(int i=0;i<5;i++){
		p2=(ListNode *)malloc(sizeof(ListNode));
		p2->val=a[i];
		p2->next=NULL;
		if(head==NULL){
			head=p1=p2;
		}else{
			p1->next=p2;
			p1=p1->next;
		}	
	}
	
	Solution sol;
	res=sol.removeNthFromEnd(head,5);
	while(res!=NULL){
    	cout<val<<"->";
        res=res->next;
    }
}