LeetCode.19.Remove Nth Node From End of List

原題鏈接：Remove Nth Node From End of List

題目內容：
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?

題目翻譯：
給定鏈表，要求一趟遍歷刪除倒數第n個元素。

Tips：設置前置指針標記結尾，與目標指針相差n位。

Python

# 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
        """
        if not head.next:
            return None
        foe = head
        current = head
        for _ in range(n):
            foe = foe.next
        if not foe:
            return head.next

        while foe.next:
            foe = foe.next
            current = current.next

        current.next = current.next.next
        return head

Python2

class Solution:
    def removeNthFromEnd(self, head, n):
        def index(node):
            if not node:
                return 0
            i = index(node.next) + 1
            if i > n:
                node.next.val = node.val
            return i
        index(head)
        return head.next