剑指 Offer II 077. 链表排序

已于 2025-03-14 14:25:32 修改
阅读量406 收藏 9
点赞数 3
分类专栏： 剑指 Offer（专项突击版） 文章标签： 归并排序
于 2025-03-14 14:25:18 首次发布
版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。
本文链接：https://blog.youkuaiyun.com/qq_42889517/article/details/145995385
版权

comments: true
edit_url: https://github.com/doocs/leetcode/edit/main/lcof2/%E5%89%91%E6%8C%87%20Offer%20II%20077.%20%E9%93%BE%E8%A1%A8%E6%8E%92%E5%BA%8F/README.md

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

    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 not head or not head.next:return head

        # 归 划分
        slow,fast=head,head.next
        while fast and fast.next:
            slow,fast=slow.next,fast.next.next
        l=head
        r=slow.next
        slow.next=None
        left=self.sortList(l)
        right=self.sortList(r)

        # 并排序
        return self.merge(left,right)

    def merge(self,left,right):
        dummy=ListNode()
        cur=dummy
        while left and right:
            if left.val<=right.val:
                cur.next=left
                left=left.next
            else:
                cur.next=right
                right=right.next
            cur=cur.next # 遗忘
        cur.next= left if not right else right
        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;
}
    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;
    }
}
    Swift
    
/** class ListNode {
*    var val: Int
*    var next: ListNode?
*    init() { self.val = 0; self.next = nil }
*    init(_ val: Int) { self.val = val; self.next = nil }
*    init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next }
* }
*/

class Solution {
    func sortList(_ head: ListNode?) -> ListNode? {
        guard let head = head, head.next != nil else {
            return head
        }

        var slow: ListNode? = head
        var fast: ListNode? = head.next

        while fast != nil && fast?.next != nil {
            slow = slow?.next
            fast = fast?.next?.next
        }

        let mid = slow?.next
        slow?.next = nil

        let left = sortList(head)
        let right = sortList(mid)

        return merge(left, right)
    }

    private func merge(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        let dummy = ListNode()
        var cur = dummy
        var l1 = l1, l2 = l2

        while let node1 = l1, let node2 = l2 {
            if node1.val <= node2.val {
                cur.next = node1
                l1 = node1.next
            } else {
                cur.next = node2
                l2 = node2.next
            }
            cur = cur.next!
        }

        cur.next = l1 ?? l2
        return dummy.next
    }
}
    评论
    添加红包

    请填写红包祝福语或标题

    个

    红包个数最小为10个

    元

    红包金额最低5元

    当前余额3.43元 前往充值 >
    需支付：10.00元
    成就一亿技术人!
    领取后你会自动成为博主和红包主的粉丝 规则
    hope_wisdom
    发出的红包
    实付元
    使用余额支付
    点击重新获取
    扫码支付
    钱包余额 0

    抵扣说明：

    1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。
    2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。

    余额充值