​【力扣 C语言】21. 合并两个有序链表​

题目

方法一 递归

思路：递归，返回较小值的节点

关键：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

/**
 * @brief  merge two lists
 * @param[in] l1
 * @param[in] l2
 * @return the node to be merged
 */
struct ListNode *mergeTwoLists(struct ListNode *l1, struct ListNode *l2)
{
    // 返回较小值的节点
    if (l1 == NULL) {
        return l2;
    } else if (l2 == NULL) {
        return l1;
    } else if (l1->val < l2->val) {
        l1->next = mergeTwoLists(l1->next, l2);
        return l1;
    } else {
        l2->next = mergeTwoLists(l1, l2->next);
        return l2;
    }
}

方法二 链表

思路：使用哑结点，简单判断并连接

关键：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

/**
 * @brief  merge two lists
 * @param[in] l1
 * @param[in] l2
 * @return the node to be merged
 */
struct ListNode *mergeTwoLists(struct ListNode *l1, struct ListNode *l2)
{
    if (l1 == NULL) {
        return l2;
    } else if (l2 == NULL) {
        return l1;
    }

    // 1. 哑结点, used to points to the head node
    struct ListNode *dummy = (struct ListNode *)malloc(sizeof(struct ListNode));
    struct ListNode *temp = dummy;
    while (l1 != NULL && l2 != NULL) {
        if (l1->val <= l2->val) {
            temp->next = l1;
            l1 = l1->next;
        } else {
            temp->next = l2;
            l2 = l2->next;
        }
        temp = temp->next;
    }
    if (l1 != NULL) {
        temp->next = l1;
    }
    if (l2 != NULL) {
        temp->next = l2;
    }

    temp = dummy->next;
    free(dummy);
    return temp;
}