Question

2046. 给按照绝对值排序的链表排序

English Version

题目描述

给你一个链表的头结点 head ，这个链表是根据结点的绝对值进行升序排序, 返回重新根据节点的值进行升序排序的链表。

 

示例 1:

输入: head = [0,2,-5,5,10,-10]
输出: [-10,-5,0,2,5,10]
解释:
根据结点的值的绝对值排序的链表是 [0,2,-5,5,10,-10].
根据结点的值排序的链表是 [-10,-5,0,2,5,10].

示例 2：

输入: head = [0,1,2]
输出: [0,1,2]
解释:
这个链表已经是升序的了。

示例 3：

输入: head = [1]
输出: [1]
解释:
这个链表已经是升序的了。

 

提示:

• 链表节点数的范围是 [1, 105].
• -5000 <= Node.val <= 5000
• head 是根据结点绝对值升序排列好的链表.

 

进阶:

• 你可以在O(n)的时间复杂度之内解决这个问题吗?

解法

方法一：头插法

先默认第一个点已经排序完毕。然后从第二个点开始，遇到值为负数的节点，采用头插法；非负数，则继续往下遍历即可。

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, val=0, next=None):
#     self.val = val
#     self.next = next
class Solution:
  def sortLinkedList(self, head: Optional[ListNode]) -> Optional[ListNode]:
    prev, curr = head, head.next
    while curr:
      if curr.val < 0:
        t = curr.next
        prev.next = t
        curr.next = head
        head = curr
        curr = t
      else:
        prev, curr = curr, curr.next
    return head

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   int val;
 *   ListNode next;
 *   ListNode() {}
 *   ListNode(int val) { this.val = val; }
 *   ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
  public ListNode sortLinkedList(ListNode head) {
    ListNode prev = head, curr = head.next;
    while (curr != null) {
      if (curr.val < 0) {
        ListNode t = curr.next;
        prev.next = t;
        curr.next = head;
        head = curr;
        curr = t;
      } else {
        prev = curr;
        curr = curr.next;
      }
    }
    return head;
  }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *   int val;
 *   ListNode *next;
 *   ListNode() : val(0), next(nullptr) {}
 *   ListNode(int x) : val(x), next(nullptr) {}
 *   ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
  ListNode* sortLinkedList(ListNode* head) {
    ListNode* prev = head;
    ListNode* curr = head->next;
    while (curr) {
      if (curr->val < 0) {
        auto t = curr->next;
        prev->next = t;
        curr->next = head;
        head = curr;
        curr = t;
      } else {
        prev = curr;
        curr = curr->next;
      }
    }
    return head;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func sortLinkedList(head *ListNode) *ListNode {
  prev, curr := head, head.Next
  for curr != nil {
    if curr.Val < 0 {
      t := curr.Next
      prev.Next = t
      curr.Next = head
      head = curr
      curr = t
    } else {
      prev, curr = curr, curr.Next
    }
  }
  return head
}