LeetCode刷题笔记(链表):remove-nth-node-from-end-of-list



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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.

解题思路

删除倒数第n个结点,我们就需要找到倒数第n个结点。于是,我们可以利用快慢指针来实现。

首先,fast走n步指向第n个结点;接着,fast和slow一起走,直到fast指向尾结点;最后,删除元素(这里需要注意的是加入pre来判断删除元素是否为头结点)。

C++版代码实现

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/**
* 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) {
if(head == NULL)
return NULL;

ListNode *slow = head;
ListNode *fast = head;
ListNode *pre = NULL;
//fast先走n步,到达第n个结点
while(--n)
fast = fast->next;
//fast和slow一起走,直到fast走到链表尾部
while(fast->next != NULL){
pre = slow;
slow = slow->next;
fast = fast->next;
}
//此处用于判断删除的结点是否为头结点
if(pre != NULL)
pre->next = slow->next;
else
head = head->next;

return head;
}
};

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完的汪(∪。∪)。。。zzz

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