LCR 077-排序链表

Raphael Liu Lv10
  2023-09-13 00:47:26 2023-09-13 00:47:26 Created   2023-09-13 16:06:28 2023-09-13 16:06:28 Updated    

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.com/problems/sort-list/

前言

这道题考虑时间复杂度优于 O(n^2) 的排序算法。题目的进阶问题要求达到 O(n \log n) 的时间复杂度和 O(1) 的空间复杂度，时间复杂度是 O(n \log n) 的排序算法包括归并排序、堆排序和快速排序（快速排序的最差时间复杂度是 O(n^2)），其中最适合链表的排序算法是归并排序。

归并排序基于分治算法。最容易想到的实现方式是自顶向下的递归实现，考虑到递归调用的栈空间，自顶向下归并排序的空间复杂度是 O(\log n)。如果要达到 O(1) 的空间复杂度，则需要使用自底向上的实现方式。

方法一：自顶向下归并排序

对链表自顶向下归并排序的过程如下。

  1. 找到链表的中点，以中点为分界，将链表拆分成两个子链表。寻找链表的中点可以使用快慢指针的做法，快指针每次移动 2 步，慢指针每次移动 1 步，当快指针到达链表末尾时，慢指针指向的链表节点即为链表的中点。

  2. 对两个子链表分别排序。

  3. 将两个排序后的子链表合并，得到完整的排序后的链表。

上述过程可以通过递归实现。递归的终止条件是链表的节点个数小于或等于 1，即当链表为空或者链表只包含 1 个节点时，不需要对链表进行拆分和排序。

[sol1-Java]
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class Solution {
    public ListNode sortList(ListNode head) {
        return sortList(head, null);
    }

    public ListNode sortList(ListNode head, ListNode tail) {
        if (head == null) {
            return head;
        }
        if (head.next == tail) {
            head.next = null;
            return head;
        }
        ListNode slow = head, fast = head;
        while (fast != tail) {
            slow = slow.next;
            fast = fast.next;
            if (fast != tail) {
                fast = fast.next;
            }
        }
        ListNode mid = slow;
        ListNode list1 = sortList(head, mid);
        ListNode list2 = sortList(mid, tail);
        ListNode sorted = merge(list1, list2);
        return sorted;
    }

    public ListNode merge(ListNode head1, ListNode head2) {
        ListNode dummyHead = new ListNode(0);
        ListNode temp = dummyHead, temp1 = head1, temp2 = head2;
        while (temp1 != null && temp2 != null) {
            if (temp1.val <= temp2.val) {
                temp.next = temp1;
                temp1 = temp1.next;
            } else {
                temp.next = temp2;
                temp2 = temp2.next;
            }
            temp = temp.next;
        }
        if (temp1 != null) {
            temp.next = temp1;
        } else if (temp2 != null) {
            temp.next = temp2;
        }
        return dummyHead.next;
    }
}
[sol1-C++]
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class Solution {
public:
    ListNode* sortList(ListNode* head) {
        return sortList(head, nullptr);
    }

    ListNode* sortList(ListNode* head, ListNode* tail) {
        if (head == nullptr) {
            return head;
        }
        if (head->next == tail) {
            head->next = nullptr;
            return head;
        }
        ListNode* slow = head, *fast = head;
        while (fast != tail) {
            slow = slow->next;
            fast = fast->next;
            if (fast != tail) {
                fast = fast->next;
            }
        }
        ListNode* mid = slow;
        return merge(sortList(head, mid), sortList(mid, tail));
    }

    ListNode* merge(ListNode* head1, ListNode* head2) {
        ListNode* dummyHead = new ListNode(0);
        ListNode* temp = dummyHead, *temp1 = head1, *temp2 = head2;
        while (temp1 != nullptr && temp2 != nullptr) {
            if (temp1->val <= temp2->val) {
                temp->next = temp1;
                temp1 = temp1->next;
            } else {
                temp->next = temp2;
                temp2 = temp2->next;
            }
            temp = temp->next;
        }
        if (temp1 != nullptr) {
            temp->next = temp1;
        } else if (temp2 != nullptr) {
            temp->next = temp2;
        }
        return dummyHead->next;
    }
};
[sol1-JavaScript]
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const merge = (head1, head2) => {
    const dummyHead = new ListNode(0);
    let temp = dummyHead, temp1 = head1, temp2 = head2;
    while (temp1 !== null && temp2 !== null) {
        if (temp1.val <= temp2.val) {
            temp.next = temp1;
            temp1 = temp1.next;
        } else {
            temp.next = temp2;
            temp2 = temp2.next;
        }
        temp = temp.next;
    }
    if (temp1 !== null) {
        temp.next = temp1;
    } else if (temp2 !== null) {
        temp.next = temp2;
    }
    return dummyHead.next;
}

const toSortList = (head, tail) => {
    if (head === null) {
        return head;
    }
    if (head.next === tail) {
        head.next = null;
        return head;
    }
    let slow = head, fast = head;
    while (fast !== tail) {
        slow = slow.next;
        fast = fast.next;
        if (fast !== tail) {
            fast = fast.next;
        }
    }
    const mid = slow;
    return merge(toSortList(head, mid), toSortList(mid, tail));
}

var sortList = function(head) {
    return toSortList(head, null);
};
[sol1-Python3]
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class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        def sortFunc(head: ListNode, tail: ListNode) -> ListNode:
            if not head:
                return head
            if head.next == tail:
                head.next = None
                return head
            slow = fast = head
            while fast != tail:
                slow = slow.next
                fast = fast.next
                if fast != tail:
                    fast = fast.next
            mid = slow
            return merge(sortFunc(head, mid), sortFunc(mid, tail))
            
        def merge(head1: ListNode, head2: ListNode) -> ListNode:
            dummyHead = ListNode(0)
            temp, temp1, temp2 = dummyHead, head1, head2
            while temp1 and temp2:
                if temp1.val <= temp2.val:
                    temp.next = temp1
                    temp1 = temp1.next
                else:
                    temp.next = temp2
                    temp2 = temp2.next
                temp = temp.next
            if temp1:
                temp.next = temp1
            elif temp2:
                temp.next = temp2
            return dummyHead.next
        
        return sortFunc(head, None)
[sol1-Golang]
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func merge(head1, head2 *ListNode) *ListNode {
    dummyHead := &ListNode{}
    temp, temp1, temp2 := dummyHead, head1, head2
    for temp1 != nil && temp2 != nil {
        if temp1.Val <= temp2.Val {
            temp.Next = temp1
            temp1 = temp1.Next
        } else {
            temp.Next = temp2
            temp2 = temp2.Next
        }
        temp = temp.Next
    }
    if temp1 != nil {
        temp.Next = temp1
    } else if temp2 != nil {
        temp.Next = temp2
    }
    return dummyHead.Next
}

func sort(head, tail *ListNode) *ListNode {
    if head == nil {
        return head
    }

    if head.Next == tail {
        head.Next = nil
        return head
    }

    slow, fast := head, head
    for fast != tail {
        slow = slow.Next
        fast = fast.Next
        if fast != tail {
            fast = fast.Next
        }
    }

    mid := slow
    return merge(sort(head, mid), sort(mid, tail))
}

func sortList(head *ListNode) *ListNode {
    return sort(head, nil)
}
[sol1-C]
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struct ListNode* merge(struct ListNode* head1, struct ListNode* head2) {
    struct ListNode* dummyHead = malloc(sizeof(struct ListNode));
    dummyHead->val = 0;
    struct ListNode *temp = dummyHead, *temp1 = head1, *temp2 = head2;
    while (temp1 != NULL && temp2 != NULL) {
        if (temp1->val <= temp2->val) {
            temp->next = temp1;
            temp1 = temp1->next;
        } else {
            temp->next = temp2;
            temp2 = temp2->next;
        }
        temp = temp->next;
    }
    if (temp1 != NULL) {
        temp->next = temp1;
    } else if (temp2 != NULL) {
        temp->next = temp2;
    }
    return dummyHead->next;
}

struct ListNode* toSortList(struct ListNode* head, struct ListNode* tail) {
    if (head == NULL) {
        return head;
    }
    if (head->next == tail) {
        head->next = NULL;
        return head;
    }
    struct ListNode *slow = head, *fast = head;
    while (fast != tail) {
        slow = slow->next;
        fast = fast->next;
        if (fast != tail) {
            fast = fast->next;
        }
    }
    struct ListNode* mid = slow;
    return merge(toSortList(head, mid), toSortList(mid, tail));
}

struct ListNode* sortList(struct ListNode* head) {
    return toSortList(head, NULL);
}

复杂度分析

  • 时间复杂度：O(n \log n)，其中 n 是链表的长度。

  • 空间复杂度：O(\log n)，其中 n 是链表的长度。空间复杂度主要取决于递归调用的栈空间。

方法二：自底向上归并排序

使用自底向上的方法实现归并排序，则可以达到 O(1) 的空间复杂度。

首先求得链表的长度 length，然后将链表拆分成子链表进行合并。

具体做法如下。

  1. 用 subLength 表示每次需要排序的子链表的长度，初始时 subLength}=1。

  2. 每次将链表拆分成若干个长度为 subLength 的子链表（最后一个子链表的长度可以小于 subLength），按照每两个子链表一组进行合并，合并后即可得到若干个长度为 subLength} \times 2 的有序子链表（最后一个子链表的长度可以小于 subLength} \times 2）。

  3. 将 subLength 的值加倍，重复第 2 步，对更长的有序子链表进行合并操作，直到有序子链表的长度大于或等于 length，整个链表排序完毕。

如何保证每次合并之后得到的子链表都是有序的呢？可以通过数学归纳法证明。

  1. 初始时 subLength}=1，每个长度为 1 的子链表都是有序的。

  2. 如果每个长度为 subLength 的子链表已经有序，合并两个长度为 subLength 的有序子链表，得到长度为 subLength} \times 2 的子链表，一定也是有序的。

  3. 当最后一个子链表的长度小于 subLength 时，该子链表也是有序的，合并两个有序子链表之后得到的子链表一定也是有序的。

因此可以保证最后得到的链表是有序的。

[sol2-Java]
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class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null) {
            return head;
        }
        int length = 0;
        ListNode node = head;
        while (node != null) {
            length++;
            node = node.next;
        }
        ListNode dummyHead = new ListNode(0, head);
        for (int subLength = 1; subLength < length; subLength <<= 1) {
            ListNode prev = dummyHead, curr = dummyHead.next;
            while (curr != null) {
                ListNode head1 = curr;
                for (int i = 1; i < subLength && curr.next != null; i++) {
                    curr = curr.next;
                }
                ListNode head2 = curr.next;
                curr.next = null;
                curr = head2;
                for (int i = 1; i < subLength && curr != null && curr.next != null; i++) {
                    curr = curr.next;
                }
                ListNode next = null;
                if (curr != null) {
                    next = curr.next;
                    curr.next = null;
                }
                ListNode merged = merge(head1, head2);
                prev.next = merged;
                while (prev.next != null) {
                    prev = prev.next;
                }
                curr = next;
            }
        }
        return dummyHead.next;
    }

    public ListNode merge(ListNode head1, ListNode head2) {
        ListNode dummyHead = new ListNode(0);
        ListNode temp = dummyHead, temp1 = head1, temp2 = head2;
        while (temp1 != null && temp2 != null) {
            if (temp1.val <= temp2.val) {
                temp.next = temp1;
                temp1 = temp1.next;
            } else {
                temp.next = temp2;
                temp2 = temp2.next;
            }
            temp = temp.next;
        }
        if (temp1 != null) {
            temp.next = temp1;
        } else if (temp2 != null) {
            temp.next = temp2;
        }
        return dummyHead.next;
    }
}
[sol2-C++]
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class Solution {
public:
    ListNode* sortList(ListNode* head) {
        if (head == nullptr) {
            return head;
        }
        int length = 0;
        ListNode* node = head;
        while (node != nullptr) {
            length++;
            node = node->next;
        }
        ListNode* dummyHead = new ListNode(0, head);
        for (int subLength = 1; subLength < length; subLength <<= 1) {
            ListNode* prev = dummyHead, *curr = dummyHead->next;
            while (curr != nullptr) {
                ListNode* head1 = curr;
                for (int i = 1; i < subLength && curr->next != nullptr; i++) {
                    curr = curr->next;
                }
                ListNode* head2 = curr->next;
                curr->next = nullptr;
                curr = head2;
                for (int i = 1; i < subLength && curr != nullptr && curr->next != nullptr; i++) {
                    curr = curr->next;
                }
                ListNode* next = nullptr;
                if (curr != nullptr) {
                    next = curr->next;
                    curr->next = nullptr;
                }
                ListNode* merged = merge(head1, head2);
                prev->next = merged;
                while (prev->next != nullptr) {
                    prev = prev->next;
                }
                curr = next;
            }
        }
        return dummyHead->next;
    }

    ListNode* merge(ListNode* head1, ListNode* head2) {
        ListNode* dummyHead = new ListNode(0);
        ListNode* temp = dummyHead, *temp1 = head1, *temp2 = head2;
        while (temp1 != nullptr && temp2 != nullptr) {
            if (temp1->val <= temp2->val) {
                temp->next = temp1;
                temp1 = temp1->next;
            } else {
                temp->next = temp2;
                temp2 = temp2->next;
            }
            temp = temp->next;
        }
        if (temp1 != nullptr) {
            temp->next = temp1;
        } else if (temp2 != nullptr) {
            temp->next = temp2;
        }
        return dummyHead->next;
    }
};
[sol2-JavaScript]
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const merge = (head1, head2) => {
    const dummyHead = new ListNode(0);
    let temp = dummyHead, temp1 = head1, temp2 = head2;
    while (temp1 !== null && temp2 !== null) {
        if (temp1.val <= temp2.val) {
            temp.next = temp1;
            temp1 = temp1.next;
        } else {
            temp.next = temp2;
            temp2 = temp2.next;
        }
        temp = temp.next;
    }
    if (temp1 !== null) {
        temp.next = temp1;
    } else if (temp2 !== null) {
        temp.next = temp2;
    }
    return dummyHead.next;
}

var sortList = function(head) {
    if (head === null) {
        return head;
    }
    let length = 0;
    let node = head;
    while (node !== null) {
        length++;
        node = node.next;
    }
    const dummyHead = new ListNode(0, head);
    for (let subLength = 1; subLength < length; subLength <<= 1) {
        let prev = dummyHead, curr = dummyHead.next;
        while (curr !== null) {
            let head1 = curr;
            for (let i = 1; i < subLength && curr.next !== null; i++) {
                curr = curr.next;
            }
            let head2 = curr.next;
            curr.next = null;
            curr = head2;
            for (let i = 1; i < subLength && curr != null && curr.next !== null; i++) {
                curr = curr.next;
            }
            let next = null;
            if (curr !== null) {
                next = curr.next;
                curr.next = null;
            }
            const merged = merge(head1, head2);
            prev.next = merged;
            while (prev.next !== null) {
                prev = prev.next;
            }
            curr = next;
        }
    }
    return dummyHead.next;
};
[sol2-Python3]
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class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        def merge(head1: ListNode, head2: ListNode) -> ListNode:
            dummyHead = ListNode(0)
            temp, temp1, temp2 = dummyHead, head1, head2
            while temp1 and temp2:
                if temp1.val <= temp2.val:
                    temp.next = temp1
                    temp1 = temp1.next
                else:
                    temp.next = temp2
                    temp2 = temp2.next
                temp = temp.next
            if temp1:
                temp.next = temp1
            elif temp2:
                temp.next = temp2
            return dummyHead.next
        
        if not head:
            return head
        
        length = 0
        node = head
        while node:
            length += 1
            node = node.next
        
        dummyHead = ListNode(0, head)
        subLength = 1
        while subLength < length:
            prev, curr = dummyHead, dummyHead.next
            while curr:
                head1 = curr
                for i in range(1, subLength):
                    if curr.next:
                        curr = curr.next
                    else:
                        break
                head2 = curr.next
                curr.next = None
                curr = head2
                for i in range(1, subLength):
                    if curr and curr.next:
                        curr = curr.next
                    else:
                        break
                
                succ = None
                if curr:
                    succ = curr.next
                    curr.next = None
                
                merged = merge(head1, head2)
                prev.next = merged
                while prev.next:
                    prev = prev.next
                curr = succ
            subLength <<= 1
        
        return dummyHead.next
[sol2-Golang]
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func merge(head1, head2 *ListNode) *ListNode {
    dummyHead := &ListNode{}
    temp, temp1, temp2 := dummyHead, head1, head2
    for temp1 != nil && temp2 != nil {
        if temp1.Val <= temp2.Val {
            temp.Next = temp1
            temp1 = temp1.Next
        } else {
            temp.Next = temp2
            temp2 = temp2.Next
        }
        temp = temp.Next
    }
    if temp1 != nil {
        temp.Next = temp1
    } else if temp2 != nil {
        temp.Next = temp2
    }
    return dummyHead.Next
}

func sortList(head *ListNode) *ListNode {
    if head == nil {
        return head
    }

    length := 0
    for node := head; node != nil; node = node.Next {
        length++
    }

    dummyHead := &ListNode{Next: head}
    for subLength := 1; subLength < length; subLength <<= 1 {
        prev, cur := dummyHead, dummyHead.Next
        for cur != nil {
            head1 := cur
            for i := 1; i < subLength && cur.Next != nil; i++ {
                cur = cur.Next
            }

            head2 := cur.Next
            cur.Next = nil
            cur = head2
            for i := 1; i < subLength && cur != nil && cur.Next != nil; i++ {
                cur = cur.Next
            }

            var next *ListNode
            if cur != nil {
                next = cur.Next
                cur.Next = nil
            }

            prev.Next = merge(head1, head2)

            for prev.Next != nil {
                prev = prev.Next
            }
            cur = next
        }
    }
    return dummyHead.Next
}
[sol2-C]
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struct ListNode* merge(struct ListNode* head1, struct ListNode* head2) {
    struct ListNode* dummyHead = malloc(sizeof(struct ListNode));
    dummyHead->val = 0;
    struct ListNode *temp = dummyHead, *temp1 = head1, *temp2 = head2;
    while (temp1 != NULL && temp2 != NULL) {
        if (temp1->val <= temp2->val) {
            temp->next = temp1;
            temp1 = temp1->next;
        } else {
            temp->next = temp2;
            temp2 = temp2->next;
        }
        temp = temp->next;
    }
    if (temp1 != NULL) {
        temp->next = temp1;
    } else if (temp2 != NULL) {
        temp->next = temp2;
    }
    return dummyHead->next;
}

struct ListNode* sortList(struct ListNode* head) {
    if (head == NULL) {
        return head;
    }
    int length = 0;
    struct ListNode* node = head;
    while (node != NULL) {
        length++;
        node = node->next;
    }
    struct ListNode* dummyHead = malloc(sizeof(struct ListNode));
    dummyHead->next = head;
    for (int subLength = 1; subLength < length; subLength <<= 1) {
        struct ListNode *prev = dummyHead, *curr = dummyHead->next;
        while (curr != NULL) {
            struct ListNode* head1 = curr;
            for (int i = 1; i < subLength && curr->next != NULL; i++) {
                curr = curr->next;
            }
            struct ListNode* head2 = curr->next;
            curr->next = NULL;
            curr = head2;
            for (int i = 1; i < subLength && curr != NULL && curr->next != NULL;
                 i++) {
                curr = curr->next;
            }
            struct ListNode* next = NULL;
            if (curr != NULL) {
                next = curr->next;
                curr->next = NULL;
            }
            struct ListNode* merged = merge(head1, head2);
            prev->next = merged;
            while (prev->next != NULL) {
                prev = prev->next;
            }
            curr = next;
        }
    }
    return dummyHead->next;
}

复杂度分析

  • 时间复杂度：O(n \log n)，其中 n 是链表的长度。

  • 空间复杂度：O(1)。

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LCR 077-排序链表