Question: Remove Nth node from end of list
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.
Method 1
One pass
Time complexity O(n)
/**
* 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)
{
ListNode*current=head;
ListNode*previous;
int cout=0;
while(current!=0&&cout<n)
{
previous=current;
current=current->next;
cout++;
}
if(current==0&&cout==1)
{
ListNode*DeleteNode=head;
delete DeleteNode;
head=NULL;
}
else if (current==0)
{
ListNode*DeleteNode=head;
head=head->next;
delete DeleteNode;
}
else
{
ListNode*rear=current;
current=head;
while(rear!=0)
{
rear=rear->next;
previous=current;
current=current->next;
}
previous->next=current->next;
delete current;
}
return head;
}
};
Method 2
Two pass
Time complexity O(n)
Explanation:One pass is to determine the number of node, the other is to delete the demanded node!
/**
* 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 NodeNumber=0;
ListNode*current=head;
while(current!=0)
{
NodeNumber++;
current=current->next;
}
if(NodeNumber==1&&n>=1)
{
head=NULL;
}
else if(NodeNumber==n)
{
ListNode*DeleteNode;
DeleteNode=head;
head=head->next;
delete DeleteNode;
}
else
{ current=head;
int j=0;
ListNode*previous;
current=head;
for(current;j<NodeNumber-n;previous=current,current=current->next)
{
j++;
}
previous->next=current->next;
delete current;
}
return head;
}
};
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