题解 | #合并两个排序的链表#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 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
     */
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
        // write code here
        if (pHead1 == nullptr && pHead2 == nullptr)
            return nullptr;
        if (pHead1 != nullptr && pHead2 == nullptr)
            return pHead1;
        if (pHead1 == nullptr && pHead2 != nullptr)
            return pHead2;
        
        ListNode *head = nullptr;
        if (pHead1->val < pHead2->val) {
            head = pHead1;
            head->next = Merge(pHead1->next, pHead2);
        } else {
            head = pHead2;
            head->next = Merge(pHead1, pHead2->next);
        }
        return head;
    }
};