题解 | #合并两个排序的链表#

class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
        if (pHead1 == nullptr) return pHead2;
        if (pHead2 == nullptr) return pHead1;
        ListNode *temp3=nullptr;
        ListNode *pHead3 = new ListNode(-1);
        temp3 = pHead3;
        while (pHead1 != nullptr && pHead2 != nullptr) {
            if (pHead1->val < pHead2->val) {
                temp3->next = pHead1;
                pHead1 = pHead1 -> next;
            } else {
                temp3->next = pHead2;
                pHead2 = pHead2 -> next;
            }
            temp3 = temp3->next;
        }
        if (pHead1 == nullptr) temp3->next = pHead2;
        else if (pHead2 == nullptr) temp3->next = pHead1;
        return pHead3->next;
    }
};

C++非递归解法，通过假头节点减少代码量。