[LeetCode] Remove Nth Node From End of List-优快云博客

本文链接：https://blog.youkuaiyun.com/a83610312/article/details/9531573

本文介绍了一种高效的方法来解决链表中删除倒数第N个节点的问题，通过双指针技巧实现一次遍历即可完成操作，提高了算法效率。

Given a linked list, remove the n^th 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.

每次遇到这种链表相关的题的时候我都习惯新建一个Guard作为头，这样可以简化很多判断。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
#define LN ListNode
class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
      LN* Guard=new LN(-1);
	  Guard->next=head;
	  LN* ans;
	  LN* pre=Guard;
	  LN* pDel=head;
	  LN* pFast=head;
	  if (n<=0)
		  ans=head;
	  while(pFast&&n--)
	  	pFast=pFast->next;
	  if (n>0)
		  ans=head;
	  else
	  {
		  while(pFast)
		  {
			  pFast=pFast->next;
			  pre=pre->next;
			  pDel=pDel->next;
		  }
		  pre->next=pDel->next;
		  delete pDel;
		  pDel=0;
		  ans=Guard->next;
	  }
	  delete Guard;
	  Guard=0;
	  return ans;
    }
};