问题描述
Given a linked list, remove the n-th node from the end of list and return its head.
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.
解法思路
此题关键在于链表的熟悉,实际上不是很难。先遍历链表,求得链表的长度,可以顺序确定删除元素的位置,然后利用链表的删除可以做到。这道题遇到的问题是当删除第一个元素时,我写的删除元素操作不适用,需要特殊分类出来!
C++代码
/**
* 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) {
int a=0;//链表长度
ListNode *temp=head;
while(temp!=nullptr)
{
temp=temp->next;
++a;
}
int b=a-n+1;//找出删除的位置
if(b==1) head=head->next;//要删去第一个元素,特殊情况
temp=head;
int c=0;
while(temp!=nullptr)
{
if(c==b-2)
{
temp->next=temp->next->next;//删除元素
break;
}
temp=temp->next;
++c;
}
return head;
}
};
计算结果
Runtime: 4 ms, faster than 100.00% of C++ online submissions for Remove Nth Node From End of List.
Memory Usage: 8.5 MB, less than 100.00% of C++ online submissions for Remove Nth Node From End of List.