【leetcode题解】【E】【61.5】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.

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最好是一次遍历就搞定，那么，可以先让一个fast指针，走n下，然后两个指针一块走，fast走到头，那么，从头开始走的指针就到了应该到的位置

# 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):

        i = 0
        fast = head.next
        while i < n:
            if not fast:
                return head.next
            fast = fast.next
            i+=1
        p = head

        while fast:
            p = p.next
            fast = fast.next

        if p.next is None:
            p.next = None
        else:
            p.next = p.next.next

        return head
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """