[LeetCode]19. Remove Nth Node From End of List

最新推荐文章于 2024-01-12 15:01:11 发布

原创最新推荐文章于 2024-01-12 15:01:11 发布 · 136 阅读

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本文介绍了一种高效的方法来移除单链表中从尾部开始的第N个节点，并提供了一个C++实现示例。通过将链表节点存储在一个数组中，该方法能够在遍历一次链表后直接定位到待删除节点并进行操作。

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.

Personal tips:用数组储存单链表顺序，然后根据下标删除，代码如下：

class Solution {
public:
	ListNode* removeNthFromEnd(ListNode* head, int n) {
		if (n == 0) return head;
		vector<ListNode*> Nth;
		ListNode *h = head;
		while (h!=NULL)
		{
			Nth.push_back(h);
			h = h->next;
		}
		if (Nth.size() == n) return head->next;
		Nth.push_back(NULL);	
		Nth[Nth.size() - 2 - n]->next = Nth[Nth.size() - n];
		return head;
	}
};