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: 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* 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