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
        ListNode* dummy=new ListNode(-1);
        ListNode* cur=dummy;
        while(pHead1&&pHead2){
            if(pHead1->val<pHead2->val){
                cur->next=new ListNode(pHead1->val);
                cur=cur->next;
                pHead1=pHead1->next;
            }
            else{
                cur->next=new ListNode(pHead2->val);
                cur=cur->next;
                pHead2=pHead2->next;
            }
        }
        while(pHead1){
            cur->next=new ListNode(pHead1->val);
            cur=cur->next;
            pHead1=pHead1->next;
        }
        while(pHead2){
            cur->next=new ListNode(pHead2->val);
            cur=cur->next;
            pHead2=pHead2->next;
        }
        return dummy->next;
    }
};