LeetCode-19. 删除链表的倒数第 N 个结点【链表双指针】

news2024/10/7 1:50:30

LeetCode-19. 删除链表的倒数第 N 个结点【链表双指针】

题目描述：
解题思路一：双指针
解题思路二：优化
解题思路三：0

题目描述：

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

示例 1：
在这里插入图片描述
输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]
示例 2：

输入：head = [1], n = 1
输出：[]
示例 3：

输入：head = [1,2], n = 1
输出：[1]

提示：

链表中结点的数目为 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

进阶：你能尝试使用一趟扫描实现吗？

解题思路一：双指针

# 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: Optional[ListNode], n: int) -> Optional[ListNode]:
        dummy_head = ListNode()
        dummy_head.next = head
        cur = pre = dummy_head
        for _ in range(n):
            cur = cur.next
        while cur.next:
            cur = cur.next
            pre = pre.next
        pre.next = pre.next.next
        return dummy_head.next

时间复杂度：O(n)
空间复杂度：O(1)

解题思路二：优化

# 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: Optional[ListNode], n: int) -> Optional[ListNode]:
        dummyHead = ListNode(next = head)
        slow, fast = dummyHead, dummyHead
        while n > 0 and fast != None:
            fast = fast.next
            n -= 1
        fast = fast.next
        while fast != None:
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return dummyHead.next