leetcode-19 Remove Nth Node From End of List (移除第N个节点)

原题：

​ Given a linked list, remove the n-th node from the end of list and return its head.

​ 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.

理解：

• 考察无头单链表移除节点代码编写

思路：

• 遍历数组得到长度
• 遍历到剔除节点的前一个节点
• 剔除节点
• 特殊情况分析
• 返回结果

代码：

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

class Solution:
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        
        len_head=1
        flag=1
        next_node=head
        
        #得到链表长度
        while next_node.next:
            next_node=next_node.next
            len_head=len_head+1
            
        #特殊情况，移除第一个节点    
        if n==len_head:
            return head.next
            
        #遍历到移除节点前一个节点
        first_node=head
        while flag

结果：

Runtime: 40 ms, faster than 99.22% of Python3 online submissions for Remove Nth Node From End of List.