Remove Nth Node From End of List

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.

好吧，整个程序分成三部分：

1. 设定一个指针先走n步，然后记录位置
2. 再来一个指针从头开始，两个指针齐头并进
3. 第一个指针找到链表的末尾，第二个指针所处地位之就是所要删除的元素。

# 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
        """
        # 初始化
        dummy = ListNode(0)
        dummy.next = head
        pre, cur = dummy, head
        # 完成步骤1
        for i in range(n-1):
            cur = cur.next
        # 完成步骤2
        while cur != None and cur.next != None:
            cur = cur.next
            pre = pre.next
        pre.next = pre.next.next
        # 完成步骤 3
        while pre != None and pre.next != None:
            pre = pre.next
        return dummy.next

上面的算法删除步骤略麻烦，来个更简洁的：

# 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
        """
        # 初始化
        fast = slow = head
        # 步骤1
        for _ in range(n):
            fast = fast.next
        # 判断是否越界
        if not fast:
            return head.next
        # 步骤2
        while fast.next:
            fast = fast.next
            slow = slow.next
        # 步骤3
        slow.next = slow.next.next
        return head