Problem:

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:

采用游标的方式，一个游标s在前边n的位置，另一个p在头部，当s走到链表结尾的时候p的位置就是应该删除的地方，时间复杂度O(n)

题目大意：

给一个单链表，要求删除倒数第n个数。

Java源代码（308ms）：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    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语言源代码（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++源代码（8ms）：

/**
 * 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 *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源代码（67ms）：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class 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