Sort List

描述

Sort a linked list in O(n log n) time using constant space complexity.

分析

常数空间且O(nlogn)，单链表适合用归并排序，双向链表适合用快速排序。本题可以复用 Merge Two Sorted Lists 的代码。

代码

Java

// Sort List
// 归并排序，时间复杂度O(nlogn)，空间复杂度O(1)
public class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null)return head;

        final ListNode middle = findMiddle(head);
        final ListNode head2 = middle.next;
        middle.next = null; // 断开

        final ListNode l1 = sortList(head);  // 前半段排序
        final ListNode l2 = sortList(head2);  // 后半段排序
        return mergeTwoLists(l1, l2);
    }

    // Merge Two Sorted Lists
    private static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        for (ListNode p = dummy; l1 != null || l2 != null; p = p.next) {
            int val1 = l1 == null ? Integer.MAX_VALUE : l1.val;
            int val2 = l2 == null ? Integer.MAX_VALUE : l2.val;
            if (val1 <= val2) {
                p.next = l1;
                l1 = l1.next;
            } else {
                p.next = l2;
                l2 = l2.next;
            }
        }
        return dummy.next;
    }
    // 快慢指针找到中间节点
    private static ListNode findMiddle(ListNode head) {
        if (head == null) return null;

        ListNode slow = head;
        ListNode fast = head.next;

        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
}

C++

// Sort List
// 归并排序，时间复杂度O(nlogn)，空间复杂度O(1)
class Solution {
public:
    ListNode *sortList(ListNode *head) {
        if (head == NULL || head->next == NULL)return head;

        ListNode *middle = findMiddle(head);
        ListNode *head2 = middle->next;
        middle->next = nullptr; // 断开

        ListNode *l1 = sortList(head);  // 前半段排序
        ListNode *l2 = sortList(head2);  // 后半段排序
        return mergeTwoLists(l1, l2);
    }
private:
    // Merge Two Sorted Lists
    static ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        ListNode dummy(-1);
        for (ListNode* p = &dummy; l1 != nullptr || l2 != nullptr; p = p->next) {
            int val1 = l1 == nullptr ? INT_MAX : l1->val;
            int val2 = l2 == nullptr ? INT_MAX : l2->val;
            if (val1 <= val2) {
                p->next = l1;
                l1 = l1->next;
            } else {
                p->next = l2;
                l2 = l2->next;
            }
        }
        return dummy.next;
    }
    // 快慢指针找到中间节点
    static ListNode* findMiddle(ListNode* head) {
        if (head == nullptr) return nullptr;

        ListNode *slow = head;
        ListNode *fast = head->next;

        while (fast != nullptr && fast->next != nullptr) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow;
    }
};

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