leetcode笔记:Remove Nth Node From End of List
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一. 题目描述
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.
二. 题目分析
给出一个链表, n
是指删除倒数第n
个节点。这里的提示n
的值默认是合法的。不过其实对输入的n进行异常判断也只需要几句语句。
使用两个指针,即快/慢指针的概念,其中一个指针先走n
步,然后慢指针走,等到快指针走到结尾时,那么慢指针走到了需要删除的节点的前一个位置。
这道题主要难点是考虑边界问题,以及特殊情况(要删除的是头节点),如输入1->2->3->4
, n=4
,那么需要删除1
,此时只需将头指针head = head->next
就可以了。
三. 示例代码
#include <iostream>struct ListNode{ int value; ListNode* next; ListNode(int x): value(x), next(NULL){};};class Solution{public: ListNode *removeNthFromEnd(ListNode *head, int n) { if (head == NULL) return NULL; ListNode *fast = head; ListNode *slow = head; ListNode *temp = head; for (int i = 0; i < n ; i++) { fast = fast->next; if (fast) continue; else break; } while (fast) { fast = fast->next; temp = slow; slow = slow->next; } if (slow == head) { head = head->next; return head; } temp->next = slow->next; delete slow; return head; }};
结果:
四. 小结
实际编程中经常会遇到边界问题,不小心的错误很容易造成程序奔溃,关于指针和链表的使用技巧还需要进一步的学习。
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