【LeetCode】19. Remove Nth Node From End of List

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19. Remove Nth Node From End of List

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?

解题思路：

（1）两遍扫描法

第一遍扫描获取长度，第二遍扫描找到被删除节点的前驱节点，然后删除节点。

已经AC的代码：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        if head is None:
            return head

        count = 0
        tempHead = head
        while tempHead:
            count += 1
            tempHead = tempHead.next

        if count == 1:
            return None

        node = head
        temp = 0
        while node:
            temp += 1
            if count - n == 0:
                head = head.next
                return head
            if temp == (count - n):
                removeNode = node.next
                lastNode = removeNode.next
                node.next = lastNode
                break
            node = node.next
        return head

（2）前后指针法

通过两个指针，让第一个指针出发n步后，第二个指针继续出发。当第一个指针到达尾部后，第二个指针刚好能够到达被删除节点。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        if not head:
            return head

        if n == 1 and head.next == None:
            return head.next

        fast = head
        slow = head
        while n:
            fast = fast.next
            n -= 1
        if not fast:
            return head.next
        while fast.next:
            fast = fast.next
            slow = slow.next
        node = slow.next.next
        slow.next = node

        return head

Reference

【1】19. 删除链表的倒数第N个节点，地址：https://zhuanlan.zhihu.com/p/105905356?utm_source=wechat_session&utm_medium=social&utm_oi=743812915018104832