LeetCode --- Remove Nth Node From End of List

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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My Submitted Code

/** * 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=head;        ListNode* p2=head;        ListNode* pre=p1;        int count=0;        while(p2->next){            if(count >= (n-1)){                pre=p1;                p1=p1->next;            }            p2=p2->next;            ++count;        }        if(count == 0){            return NULL;        }        if(p1){            pre->next=p1->next;          }else{            pre->next=NULL;        }        if(p1== head){            return head->next;        }        return head;    }};