leetcode-19-remove nth node from end of list

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.

移除倒数第N个结点。

用双指针，让快指针先走，主要要删除的结点为首尾的情况。

/**

• 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->next&&n==1) return NULL;
      ListNode* fast = head;
      ListNode* slow = fast;
      for(int i=0;i<n;i++)
      fast=fast->next;
      if(!fast) return head->next;
      while(fast->next){
          fast=fast->next;
          slow=slow->next;
      }
      slow->next=slow->next->next;
      return head;
  
  }
  

  };