Leetcode Remove Nth Node from End of List (C,c++,java,python)

Problem:

Given A linked list, remove the nth node from the end of the list and return its head.

For example,

   1->2->3->4->5  N = 2.   1->2->3->5.

Note:
Given n would always be valid.
Try to do the in one pass.

Solution: In the way of a cursor, a cursor s in the position of the front N, another p in the head, when s go to the end of the list when the P position is the place to be deleted, time complexity O (n)
To give a single linked list, ask to delete the number of the count of the nth.
Java source Code (308MS):

/** * Definition for singly-linked list. * public class ListNode {*     int val; *     ListNode Next; *     listnode (int x) {val = x;}}} */public class Soluti On {public    ListNode removenthfromend (listnode head, int n) {        listnode s,p=new listnode (0);        S=head;p.next=head;        while (n>0 && s!=null) {            s=s.next;            n--;        }        while (s!=null) {            s=s.next;            P=p.next;        }        if (p.next==head) {            head=head.next;        } else{            p.next=p.next.next;        }        return head;    }}

C Language Source code (6MS):

/** * Definition for singly-linked list. * struct ListNode {*     int val; *     struct ListNode *next; *}; */struct listnode* removenthfromend (struct listnode* Head, int n) {    struct listnode *s,*p= (struct listnode*) malloc (sizeof (struct listnode));    s=head;p->next=head;    while (n--&& s!=null) s=s->next;    while (s!=null) {        s=s->next;        p=p->next;    }    if (p->next==head) {        head=head->next;    } else{        p->next=p->next->next;    }    return head;}

C + + source code (8MS):

/** * Definition for singly-linked list. * struct ListNode {*     int val; *     ListNode *next; *     listnode (int x): Val (x), Next (NULL) {} *}; */class Soluti On {public:    listnode* removenthfromend (listnode* head, int n) {        listnode *s,*p=new listnode (0);        s=head;p->next=head;        while (n--&& s!=null) s=s->next;        while (s!=null) {            s=s->next;            p=p->next;        }        if (p->next==head) {            head=head->next;        } else{            p->next=p->next->next;        }        return head;    };

Python source code (67MS):

# Definition for singly-linked list.# class listnode:#     def __init__ (self, x): #         self.val = x#         Self.next = Nonec Lass Solution:    # @param {ListNode} head    # @param {integer} n    # @return {listnode}    def removenthfromend ( Self, head, N):        s=head;p=listnode (0)        P.next=head while        n>0 and S!=none:            n-=1;s=s.next        While S!=none:            s=s.next;p=p.next        if P.next==head:head=head.next        else:p.next=p.next.next        Return head

Leetcode Remove Nth Node from End of List (C,c++,java,python)