19. Remove Nth Node From End of

【Description】

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.

【Idea】

• 第一种解法：
  遍历该逆向列表，并用list存储。del nums[length-n] 删除指定下标元素，再重新遍历list，创建新链表。提交结果如图，空间占用较高。

  
  一.png

• 第二种解法：
  在原链表基本上，创建两个指针fast 和slow， 对链表进行时间复杂度为O（n）的遍历，中fast所指节点的位置比slow 快n。遍历至fast.next==None,将slow.next.next赋给slow.next。最后返回head。提交结果如下（空间并没有低。。）：

  
  二.png

【Solution】

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

# solution 1
class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        temp = head
        nums = []
        while temp:
            nums.append(temp.val)
            temp = temp.next
        length = len(nums)
        if n > length:
            return 
        if length == 1 and n == 1:
            return []
        del nums[length-n]
        head = ListNode(None)
        prev = head
        for i in nums:
            curr = ListNode(i)
            prev.next = curr
            prev = prev.next
        return head.next

# solution 2

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        fast = slow = head
        for i in range(n):
            fast = fast.next
        if fast == None:
            return head.next 
        while fast.next:
            slow = slow.next
            fast = fast.next
        slow.next = slow.next.next
        return head