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leetcode-Remove Nth Node From End of List
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.
Note:
Given n will always be valid.
Try to do this in one pass.
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Linked List Two PointersHave you met this question in a real interview?
/**
* 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 *p1,*p2;
p1=head;
p2=head;
for(int i=1;i<=n;i++)
p2=p2->next;
if(p2==NULL)
{
p1=p1->next;
return p1;
}
while(p2->next!=NULL)
{
p1=p1->next;
p2=p2->next;
}
p1->next=p1->next->next;
return head;
}
};
leetcode-Remove Nth Node From End of List
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