Question

剑指 Offer II 077. 链表排序

题目描述

给定链表的头结点 head ，请将其按 升序 排列并返回 排序后的链表 。

 

示例 1：

输入：head = [4,2,1,3]
输出：[1,2,3,4]

示例 2：

输入：head = [-1,5,3,4,0]
输出：[-1,0,3,4,5]

示例 3：

输入：head = []
输出：[]

 

提示：

• 链表中节点的数目在范围 [0, 5 * 104] 内
• -105 <= Node.val <= 105

 

进阶：你可以在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序吗？

 

注意：本题与主站 148 题相同：https://leetcode.cn/problems/sort-list/

解法

方法一

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, val=0, next=None):
#     self.val = val
#     self.next = next
class Solution:
  def sortList(self, head: ListNode) -> ListNode:
    if head is None or head.next is None:
      return head
    slow, fast = head, head.next
    while fast and fast.next:
      slow, fast = slow.next, fast.next.next
    t = slow.next
    slow.next = None
    l1, l2 = self.sortList(head), self.sortList(t)
    dummy = ListNode()
    cur = dummy
    while l1 and l2:
      if l1.val <= l2.val:
        cur.next = l1
        l1 = l1.next
      else:
        cur.next = l2
        l2 = l2.next
      cur = cur.next
    cur.next = l1 or l2
    return dummy.next

/**
 * 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 sortList(ListNode head) {
    if (head == null || head.next == null) {
      return head;
    }
    ListNode slow = head, fast = head.next;
    while (fast != null && fast.next != null) {
      slow = slow.next;
      fast = fast.next.next;
    }
    ListNode t = slow.next;
    slow.next = null;
    ListNode l1 = sortList(head);
    ListNode l2 = sortList(t);
    ListNode dummy = new ListNode();
    ListNode cur = dummy;
    while (l1 != null && l2 != null) {
      if (l1.val <= l2.val) {
        cur.next = l1;
        l1 = l1.next;
      } else {
        cur.next = l2;
        l2 = l2.next;
      }
      cur = cur.next;
    }
    cur.next = l1 == null ? l2 : l1;
    return dummy.next;
  }
}

/**
 * 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* sortList(ListNode* head) {
    if (!head || !head->next) return head;
    auto* slow = head;
    auto* fast = head->next;
    while (fast && fast->next) {
      slow = slow->next;
      fast = fast->next->next;
    }
    auto* t = slow->next;
    slow->next = nullptr;
    auto* l1 = sortList(head);
    auto* l2 = sortList(t);
    auto* dummy = new ListNode();
    auto* cur = dummy;
    while (l1 && l2) {
      if (l1->val <= l2->val) {
        cur->next = l1;
        l1 = l1->next;
      } else {
        cur->next = l2;
        l2 = l2->next;
      }
      cur = cur->next;
    }
    cur->next = l1 ? l1 : l2;
    return dummy->next;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func sortList(head *ListNode) *ListNode {
  if head == nil || head.Next == nil {
    return head
  }
  slow, fast := head, head.Next
  for fast != nil && fast.Next != nil {
    slow, fast = slow.Next, fast.Next.Next
  }
  t := slow.Next
  slow.Next = nil
  l1, l2 := sortList(head), sortList(t)
  dummy := &ListNode{}
  cur := dummy
  for l1 != nil && l2 != nil {
    if l1.Val <= l2.Val {
      cur.Next = l1
      l1 = l1.Next
    } else {
      cur.Next = l2
      l2 = l2.Next
    }
    cur = cur.Next
  }
  if l1 != nil {
    cur.Next = l1
  } else {
    cur.Next = l2
  }
  return dummy.Next
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *   val: number
 *   next: ListNode | null
 *   constructor(val?: number, next?: ListNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 *   }
 * }
 */

function sortList(head: ListNode | null): ListNode | null {
  if (head == null || head.next == null) return head;
  // 快慢指针定位中点
  let slow: ListNode = head,
    fast: ListNode = head.next;
  while (fast != null && fast.next != null) {
    slow = slow.next;
    fast = fast.next.next;
  }
  // 归并排序
  let mid: ListNode = slow.next;
  slow.next = null;
  let l1: ListNode = sortList(head);
  let l2: ListNode = sortList(mid);
  let dummy: ListNode = new ListNode();
  let cur: ListNode = dummy;
  while (l1 != null && l2 != null) {
    if (l1.val <= l2.val) {
      cur.next = l1;
      l1 = l1.next;
    } else {
      cur.next = l2;
      l2 = l2.next;
    }
    cur = cur.next;
  }
  cur.next = l1 == null ? l2 : l1;
  return dummy.next;
}

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var sortList = function (head) {
  if (!head || !head.next) {
    return head;
  }
  let slow = head;
  let fast = head.next;
  while (fast && fast.next) {
    slow = slow.next;
    fast = fast.next.next;
  }
  let t = slow.next;
  slow.next = null;
  let l1 = sortList(head);
  let l2 = sortList(t);
  const dummy = new ListNode();
  let cur = dummy;
  while (l1 && l2) {
    if (l1.val <= l2.val) {
      cur.next = l1;
      l1 = l1.next;
    } else {
      cur.next = l2;
      l2 = l2.next;
    }
    cur = cur.next;
  }
  cur.next = l1 || l2;
  return dummy.next;
};

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   public int val;
 *   public ListNode next;
 *   public ListNode(int val=0, ListNode next=null) {
 *     this.val = val;
 *     this.next = next;
 *   }
 * }
 */
public class Solution {
  public ListNode SortList(ListNode head) {
    if (head == null || head.next == null)
    {
      return head;
    }
    ListNode slow = head, fast = head.next;
    while (fast != null && fast.next != null)
    {
      slow = slow.next;
      fast = fast.next.next;
    }
    ListNode t = slow.next;
    slow.next = null;
    ListNode l1 = SortList(head);
    ListNode l2 = SortList(t);
    ListNode dummy = new ListNode();
    ListNode cur = dummy;
    while (l1 != null && l2 != null)
    {
      if (l1.val <= l2.val)
      {
        cur.next = l1;
        l1 = l1.next;
      }
      else
      {
        cur.next = l2;
        l2 = l2.next;
      }
      cur = cur.next;
    }
    cur.next = l1 == null ? l2 : l1;
    return dummy.next;
  }
}