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Medium
Linked List
Two Pointers
Divide and Conquer
Sorting
Merge Sort

148. Sort List

中文文档

Description

Given the head of a linked list, return the list after sorting it in ascending order.

 

Example 1:

Input: head = [4,2,1,3]
Output: [1,2,3,4]

Example 2:

Input: head = [-1,5,3,4,0]
Output: [-1,0,3,4,5]

Example 3:

Input: head = []
Output: []

 

Constraints:

  • The number of nodes in the list is in the range [0, 5 * 104].
  • -105 <= Node.val <= 105

 

Follow up: Can you sort the linked list in O(n logn) time and O(1) memory (i.e. constant space)?

Solutions

Solution 1

Python3

# 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

Java

/**
 * 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;
    }
}

C++

/**
 * 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;
    }
};

Go

/**
 * 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
}

TypeScript

/**
 * 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;
}

Rust

// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn sort_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        fn merge(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
            match (l1, l2) {
                (None, Some(node)) | (Some(node), None) => Some(node),
                (Some(mut node1), Some(mut node2)) => {
                    if node1.val < node2.val {
                        node1.next = merge(node1.next.take(), Some(node2));
                        Some(node1)
                    } else {
                        node2.next = merge(Some(node1), node2.next.take());
                        Some(node2)
                    }
                }
                _ => None,
            }
        }

        fn sort(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
            if head.is_none() || head.as_ref().unwrap().next.is_none() {
                return head;
            }
            let mut head = head;
            let mut length = 0;
            let mut cur = &head;
            while cur.is_some() {
                length += 1;
                cur = &cur.as_ref().unwrap().next;
            }
            let mut cur = &mut head;
            for _ in 0..length / 2 - 1 {
                cur = &mut cur.as_mut().unwrap().next;
            }
            let right = cur.as_mut().unwrap().next.take();

            merge(sort(head), sort(right))
        }
        sort(head)
    }
}

JavaScript

/**
 * 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;
};

C#

/**
 * 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;
    }
}