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) {

        int p1,p2;

        auto LL = ListNode(-1);

        ListNode* PreHead = &LL,*temp = ≪

        while (pHead1 || pHead2) {

            pHead1 != nullptr ? p1 = pHead1->val : p1 = 9999;
            pHead2 != nullptr ? p2 = pHead2->val : p2 = 9999;

            if( p1 <= p2 ){
                temp->next = new ListNode(p1);
                pHead1 = pHead1->next;
            }else{
                temp->next = new ListNode(p2);
                pHead2 = pHead2->next;
            }
            temp = temp->next;
        }

        return PreHead->next;

    }
};