C语言 | Leetcode C语言题解之第148题排序链表

题目：

题解：

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;
}