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.
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
if(head==NULL){
return NULL;
}
ListNode* dummyHead=new ListNode(0);
dummyHead->next=head;
//p->target...->q(NULL)
ListNode* p=dummyHead;
ListNode* q=dummyHead;
for(int i=0;i<n+1;i++){
//ensure q !=NULL
assert(q);
q=q->next;
}
//find node
while(q != NULL){
p=p->next;
q=q->next;
}
//delete node
ListNode* delNode=p->next;
p->next=delNode->next;
delete delNode;
ListNode* ret=dummyHead->next;
delete dummyHead;
return ret;
}
};
