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.
/**
* 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 phead(-1);
phead.next = head;
ListNode* fast = head;
ListNode* slow = head;
ListNode* pre = &phead;
for (int i = 0; i < n;i++)
fast = fast->next;
while (fast!=NULL)
{
fast = fast->next;
slow = slow->next;
pre = pre->next;
}
pre->next = slow->next;
free(slow);
return phead.next;
}
};
