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; *     struct ListNode *next; * }; */struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {    struct ListNode* p1 = head;    struct ListNode* p2 = head;        for(int i = 0; i < n; i++){        p1 = p1->next;    }    if(!p1){ //special condition: delete head        head = head->next;        free(p2);        return head;    }        while(p1->next){        p1 = p1->next;        p2 = p2->next;    }        p1 = p2->next;    p2->next = p2->next->next;    free(p1);    return head;}