[LeetCode] 085: Remove Nth Node From End of List

最新推荐文章于 2024-09-05 06:26:26 发布

1.02^365的成长裂变

最新推荐文章于 2024-09-05 06:26:26 发布

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本文介绍了一种高效的算法，用于解决从链表中删除倒数第N个节点的问题。通过一次遍历实现，并附带了C++代码示例。适用于面试准备和技术学习。

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[Problem]

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.

[Solution]

/**
 * 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) {
		// Note: The Solution object is instantiated only once and is reused by each test case.
		// invalid
		if(n <= 0 || NULL == head)return head;

		// move the first pointer n steps forward
		int i = 0;
		ListNode *p = head;
		for(i = 0; i < n && p != NULL; ++i){
			p = p->next;
		}

		// n is larger than the length of the list
		if(i != n)return head;

		// the node to be deleted is the head of the list
		if(NULL == p){
			ListNode *q = head;
			head = head->next;
			delete q;
			return head;
		}

		// get the node to be deleted
		ListNode *pre = head;
		ListNode *q = head;
		while(p != NULL){
			pre = q;
			q = q->next;
			p = p->next;
		}

		// delete the node
		pre->next = q->next;
		delete q;

		return head;
	}
};