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)
        {
            return pHead2;
        }
        else if(pHead2==nullptr)
        {
            return pHead1;
        }

        ListNode* pMergeHead=nullptr;

        if(pHead1->val<pHead2->val)
        {
            pMergeHead=pHead1;
            pMergeHead->next=Merge(pHead1->next, pHead2);
        }
        else 
        {
            pMergeHead=pHead2;
            pMergeHead->next=Merge(pHead1, pHead2->next);
        }
        return pMergeHead;
    }
};