LeetCode 19. Remove Nth Node From End of List

题目

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.

Note:
Given n will always be valid.
Follow up:
Could you do this in one pass?

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分析

其实写过C++的代码了，One pass的做法就是通过求差来巧妙的标记位置。比如倒数第三个，那么我先走3个，然后走到头，那么我后面走的个数就是倒数第三个距离头部的距离，大概意思就是这样。
40ms,方法都差不多。

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
        """
        temp = head
        k = n
        while k != 0:
            temp = temp.next
            k -= 1
        if temp == None:
            return head.next

        pre = head
        while temp != None:
            pre = pre.next
            temp = temp.next
        pre.next = pre.next.next
        return head