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.

 

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　　简单的快慢指针问题。

#include 
      
       using namespace std;/** * 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) {        if(head==NULL)  return NULL;        ListNode * fastp = head, * slowp = head;        for(int i =0;i
       
        next;        if(fastp==NULL) return head->next;        while(fastp->next!=NULL){            fastp = fastp ->next;            slowp = slowp ->next;        }        slowp->next = slowp->next->next;        return head;    }};int main(){    return 0;}