19. Remove Nth Node From End of List(C++,Python)

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

 For example,

    Given linked list: ->->->->, and n = .

    After removing the second node from the end, the linked list becomes ->->->.
 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* cur;
     struct ListNode* pre;
     cur = head;
     pre = head;
     if(head == NULL)
         return head;
     while(n)
     {
         cur = cur->next;
         n--;
     }
     if(cur == NULL){
         head = head->next;
         return head;
     }
     while(cur->next != NULL){
         cur = cur->next;
         pre = pre->next;
     }
     pre->next = pre->next->next;
     return head;

 }

python版本：

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

class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        #这里的头结点是第一个结点
        #判断链表是否为空
        if head == None:
            return head

        #设置两个标志位
        cur = head
        pre = head

        #cur先走n步
        while n:
            cur = cur.next
            n -= 1    

        #如果cur为空，则说明n为链表长度
        if cur == None:
            return head.next

        #pre标志位和cur一起走，直到cur走到最后一个结点
        while cur.next != None:
            cur = cur.next
            pre = pre.next
        #删除倒数第n个结点
        pre.next = pre.next.next
        return head