【力扣 简单 C】21. 合并两个有序链表

目录

题目

解法一：迭代

解法二：递归

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题目

解法一：迭代

struct ListNode* merge(struct ListNode* head1, struct ListNode* head2)
{
    struct ListNode* virHead = malloc(sizeof(*virHead));
    struct ListNode* preNode = virHead;
    while (head1 && head2)
    {
        if (head1->val < head2->val)
        {
            preNode->next = head1;
            head1 = head1->next;
        }
        else
        {
            preNode->next = head2;
            head2 = head2->next;
        }
        preNode = preNode->next;
    }
    preNode->next = head1 ? head1 : head2;
 
    struct ListNode* retHead = virHead->next;
    free(virHead);
    return retHead;
}
 
struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2)
{
    return merge(list1, list2);
}

解法二：递归

void swap(struct ListNode** a, struct ListNode** b)
{
    struct ListNode* tmp = *a;
    *a = *b;
    *b = tmp;
}

struct ListNode* merge(struct ListNode* head1, struct ListNode* head2)
{
    if (!head1)
        return head2;

    if (!head2)
        return head1;

    if (head1->val > head2->val)
        swap(&head1, &head2);

    head1->next = merge(head1->next, head2);
    return head1;
}

struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2)
{
    return merge(list1, list2);
}