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) { // Start typing your C/C++ solution below // DO NOT write int main() function if(head==NULL)return NULL; vector<ListNode*> record; ListNode* ptr=head; while(ptr!=NULL){ record.push_back(ptr); ptr=ptr->next; } int total=record.size(); if(record.size()==n)return head->next; else if(n==1){ record[total-2]->next=NULL; } else{ record[total-n-1]->next=record[total-n+1]; } return head; } };