148. 排序链表

题目描述

给你链表的头结点 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？

方法一

Java

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;
    }
}

C++

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;
    }
};

TypeScript

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;
}

Python

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