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.

solution: two pointers, 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 || head->next == NULL){            return NULL;        }        ListNode *a = head;        ListNode *b = head;        for(int i = 0 ; i < n; i++){            b = b->next;        }        if(b==NULL){            return a->next;        }        while(b->next != NULL){            a = a->next;            b = b->next;        }        //del a->next.        ListNode *c = a->next->next;        a->next = c;        return head;    }};