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*dfs(ListNode* pHead1, ListNode* pHead2)
     {
        if(pHead1==nullptr || pHead2==nullptr)
            return pHead1==nullptr ? pHead2:pHead1;
        //重复子问题
        if(pHead1->val < pHead2->val)
        {
            pHead1->next = Merge(pHead1->next,pHead2);
            return pHead1;
        }
       else
        {
            pHead2->next = Merge(pHead2->next,pHead1);
            return pHead2;
        }
        return pHead1==nullptr ?pHead2:pHead1;
     }
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) 
    {
        // write code here
        //出口
        return dfs(pHead1,pHead2);
    }
};