题解 | #合并k个已排序的链表#

/**
 * struct ListNode {
 *  int val;
 *  struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param lists ListNode类一维数组
 * @param listsLen int lists数组长度
 * @return ListNode类
 */

struct ListNode *Merge(struct ListNode *pHead1, struct ListNode *pHead2)
{
    if (pHead1 == NULL && pHead2 == NULL)
        return NULL;
    struct ListNode *res = (struct ListNode *)malloc(sizeof(struct ListNode) * 1);
    struct ListNode *dumy = res;
    struct ListNode *l1 = pHead1;
    struct ListNode *l2 = pHead2;
    while (l1 != NULL && l2 != NULL)
    {
        struct ListNode *node = (struct ListNode *)malloc(sizeof(struct ListNode) * 1);
        if (l1->val < l2->val)
        {
            node->val = l1->val;
            l1 = l1->next;
        }
        else
        {
            node->val = l2->val;
            l2 = l2->next;
        }
        res->next = node;
        res = res->next;
    }
    res->next = l1 == NULL ? l2 : l1;
    return dumy->next;
}

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param lists ListNode类一维数组 输入：[{1,2},{1,4,5},{6}]
 * @param listsLen int lists数组长度
 * @return ListNode类
 */
struct ListNode *mergeKLists(struct ListNode **lists, int listsLen)
{
    struct ListNode *result = lists[0];
    for (int i = 1; i < listsLen; i++)
        result = Merge(result, lists[i]);
    return result;
}