题解 | #合并两个排序的链表#

1. 有至少一个链表为空

        if (pHead1 == NULL) return pHead2;
        if (pHead2 == NULL) return pHead1;

2. 定义虚拟头节点，存放结果

        ListNode* result = new ListNode(0);
        result->next = pHead1;
        ListNode* cur = result;

3. 当节点不为空时，比较两个节点值大小

        while (p1 != NULL && p2 != NULL) {
            if (p1->val < p2->val) {
                cur->next = p1;
                cur = cur->next;
                p1 = p1->next;
            }
            else {
                cur->next = p2;
                cur = cur->next;
                p2 = p2->next;
            }
        }

4. 有残余项时

        if (p1 != NULL && p2 == NULL) cur->next = p1;
        if (p1 == NULL && p2 != NULL) cur->next = p2;