leetcode: Remove Nth Node From End of List

最新推荐文章于 2024-09-05 06:26:26 发布

原创最新推荐文章于 2024-09-05 06:26:26 发布 · 546 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #链表 #C++ #algorithm

C/C++ 同时被 2 个专栏收录

33 篇文章

订阅专栏

Algorithm

29 篇文章

订阅专栏

本文介绍了一种算法，该算法可以在一次遍历中移除链表中的倒数第N个节点，并返回修改后的链表头。通过将所有节点指针存储在一个向量中来实现这一目标。

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.

思路

把所有的指针全部存下来，想不到其他的只遍历一次的方法了。自己中间马虎，提交了好几次才成功。一定警告自己注意边界条件。。

/**
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        vector<ListNode *> pt;
		ListNode * cur = head;
		while(cur != NULL)
		{
			pt.push_back(cur);
			cur = cur->next;
		}

		if(pt.size() == 0) return head;
		if(n <= 0 || n > pt.size()) return head;
		if(pt.size() == 1)
		{
			delete head;
			return NULL;
		}
		if(pt.size() == n)
		{
			cur = head->next;
			head->next = NULL;
			delete head;
			return cur;
		}
		if(n == 1)
		{
			pt[pt.size()-2]->next = NULL;
			delete pt[pt.size() - 1];
			return head;
		}
		ListNode * pr = pt[pt.size() - n - 1];
		cur = pr->next;
		pr->next = cur->next;
		cur->next = NULL;
		delete cur;
		return head;
    }
};