排序链表

合集 - LeetCode算法初级(6)

1.删除排序数组中的重复项2023-05-112.环状替换法详解2023-05-163.哈希表简单应用—两数之和2023-05-174.重复的子字符串问题2023-07-205.买卖股票的最佳时机 II2023-05-11

6.排序链表2024-04-01

收起

1、非递归

// 递归的归并排序
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 *= 2) {
            ListNode *prev = dummyHead, *curr = dummyHead->next;
            while (curr != nullptr) {
                ListNode* head1 = curr;
                // 取子长度的链表
                for (int i = 1; i < subLength && curr->next != nullptr; ++i)
                    curr = curr->next;
                ListNode* head2 = curr->next;
                curr->next = nullptr;
                curr = head2;
                if (curr == nullptr) {
                    ListNode* merged = merge(head1, head2);
                    prev->next = merged;
                    break;
                }
                // 不只是取第一个和第二个链表的头节点，需要记录第二个链表的末尾，不然后面很难遍历
                for (int i = 1; i < subLength && curr->next != nullptr; ++i) {
                    curr = curr->next;
                }
                // 下一次遍历从下一个结尾
                ListNode* next = nullptr;
                if (curr->next != 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;
    }
    // 两两合并
    ListNode* merge(ListNode* head1, ListNode* head2) {
        ListNode* dummmy = new ListNode();
        ListNode* p = dummmy;
        while (head1 && head2) {
            if (head1->val < head2->val) {
                p->next = head1;
                head1 = head1->next;
            } else {
                p->next = head2;
                head2 = head2->next;
            }
            p = p->next;
        }
        if (head1) {
            p->next = head1;
        } else {
            p->next = head2;
        }
        p = dummmy->next;
        delete dummmy;
        return p;
    }
};

2、递归

//递归的归并排序
class Solution {
public:
    ListNode* sortList(ListNode* head) {
        return sortList(head, nullptr);
    }
    ListNode* sortList(ListNode* head1, ListNode* head2)
    {
        //从节点head1到节点head2进行排序
        if (head1 == nullptr)
            return head1;
        if (head1->next == head2)
        {
            head1->next = nullptr; //因为从begin,mid) [mid,last]
            return head1;
        }
        //取终点再次进行折半
        ListNode* fast = head1, * slow = head1;
        while (fast != head2)
        {
            fast = fast->next;
            slow = slow->next;
            if (fast != head2)
                fast = fast->next;
        }
        ListNode* l1 = sortList(head1, slow);
        ListNode* l2 = sortList(slow, head2);
        return merge(l1, l2);
    }
    //两两合并
    ListNode* merge(ListNode* head1, ListNode* head2)
    {
        ListNode* dummmy = new ListNode();
        ListNode* p = dummmy;
        while (head1 && head2)
        {
            if (head1->val < head2->val)
            {
                p->next = head1;
                head1 = head1->next;
            }
            else
            {
                p->next = head2;
                head2 = head2->next;
            }
            p = p->next;
        }
        if (head1)
            p->next = head1;
        else
            p->next = head2;
        p = dummmy->next;
        delete dummmy;
        return p;
    }
};