LeetCode 19 - Remove Nth Node From End of List
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一、问题描述
Description: 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
.
给一个单链表,以及一个正整数 n,移除链表倒数第 n 个结点。
二、解题报告
要移除某个结点,我们只需要找到它的前一个结点即可。这里使用 快慢指针 找到倒数第 n 个结点的前一个结点,然后移除即可。
struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {}};
为了方便,这里创建了一个”头指针”,代码如下:
class Solution {public: ListNode* removeNthFromEnd(ListNode* head, int n) { ListNode* p = new ListNode(0); p->next = head; ListNode* slow = p; // 快慢指针 ListNode* fast = p; while(n--) // fast先行n步 fast = fast->next; while(fast->next != NULL) { slow = slow->next; fast = fast->next; } slow->next = slow->next->next; return p->next; }};
AC 时间 4ms.
LeetCode答案源代码:https://github.com/SongLee24/LeetCode
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