LeetCode 19 [Remove Nth Node From End of List]

原题

给定一个链表，删除链表中倒数第n个节点，返回链表的头节点。
注意事项
链表中的节点个数大于等于n

样例
给出链表**1->2->3->4->5->null
和 n = 2.
删除倒数第二个节点之后，这个链表将变成1->2->3->5->null.
**

解题思路

• 链表问题 - 快慢指针解决(Two Pointers)
• 块指针先走n步，然后快慢指针一起走，快指针走到尾，慢指针距离尾部差n个节点
• 删除节点 node.next = node.next.next

完整代码

# 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
        """
        slow, fast = head, head
        for i in range(n):
            fast = fast.next
            
        if not fast:
            head = slow.next
        else:
            while fast.next != None:
                fast = fast.next
                slow = slow.next
            slow.next = slow.next.next
        
        return head