题目地址(143. 重排链表)

题目地址(143. 重排链表)

https://leetcode-cn.com/problems/reorder-list/

题目描述

给定一个单链表 L 的头节点 head ，单链表 L 表示为：

L0 → L1 → … → Ln - 1 → Ln


请将其重新排列后变为：

L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …

不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

 

示例 1：

输入：head = [1,2,3,4]
输出：[1,4,2,3]

示例 2：

输入：head = [1,2,3,4,5]
输出：[1,5,2,4,3]

 

提示：

链表的长度范围为 [1, 5 * 104]
1 <= node.val <= 1000

前置知识

公司

• 暂无

思路

关键点

代码

解法一

• 语言支持：Python3

Python3 Code:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reorderList(self, head: ListNode) -> None:
        """
        Do not return anything, modify head in-place instead.
        """
        # 使用线性表
        if not head:
            return 
        
        vec = list()
        node = head 
        while node:
            vec.append(node)
            node = node.next 
        
        i,j = 0,len(vec)-1
        while i<j:
            vec[i].next = vec[j]
            i += 1
            # 判断是否两个节点是否相遇
            if i == j:
                break
            vec[j].next = vec[i]
            j -= 1
        # 重排 需要断开连接
        vec[i].next = None 
        
        

复杂度分析

令 n 为数组长度。

• 时间复杂度：$O(n)$
• 空间复杂度：$O(n)$

解法二

• 语言支持：Python3

Python3 Code:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reorderList(self, head: ListNode) -> None:
        """
        Do not return anything, modify head in-place instead.
        """
        # 寻找链表中点 + 反转右端链表 + 合并链表
        def findCenter(head:ListNode)->ListNode:
            slow = fast = head 
            while fast.next and fast.next.next:
                slow = slow.next 
                fast = fast.next.next 
            return slow 
        
        def reverse(head:ListNode)->ListNode:
            pre = None 
            cur = head
            while cur:
                next = cur.next 
                cur.next = pre
                pre = cur 
                cur = next 
            return pre 
        
        def merge(head1:ListNode,head2:ListNode)->ListNode:
            while head1 and head2:
                l1 = head1.next 
                l2 = head2.next 

                head1.next = head2
                head1 = l1 

                head2.next = head1
                head2 = l2 
        
        pre = findCenter(head)
        head2 = reverse(pre.next)
        pre.next = None 
        merge(head,head2)
            


        
        

复杂度分析

令 n 为数组长度。

• 时间复杂度：$O(n)$
• 空间复杂度：$O(1)$