19. Remove Nth Node From End of List

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. 

解析

  • 两种思路
  • 思路一:第一遍遍历得到链表的长度,第二遍走cnt-n步,注意边界条件
  • 思路二:两个指针,第一个先走n-1步,然后两个指针一起走

// 19. Remove Nth Node From End of List
class Solution {
public:
	ListNode* removeNthFromEnd(ListNode* head, int n) {

		ListNode* cur = head;
		if (head==NULL)
		{
			return NULL;
		}
		int cnt = 0;
		while (cur!=NULL)
		{
			cnt++;
			cur = cur->next;
		}
		cur = head;
		int pos = cnt - n+1;
		if (pos==1)
		{
			return cur->next;
		}
		pos = pos - 2;
		while (pos)
		{
			cur = cur->next;
			pos--;
		}
		ListNode* dete = cur->next;
		if (dete->next)
		{
			cur->next = cur->next->next;
		}
		else
		{
			cur->next = NULL;
		}
		
		
		//delete dete;
		return head;
	}
};

题目来源

原文地址:https://www.cnblogs.com/ranjiewen/p/8324367.html