题解 | #合并两个排序的链表#

/**
 * 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 pHead1 ListNode类
 * @param pHead2 ListNode类
 * @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;
}