题解 | #合并两个排序的链表#双指针交替运行

/**
 * 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) {
        if (pHead1 == nullptr) return pHead2;
        if (pHead2 == nullptr) return pHead1;
        if (!pHead1 && !pHead2) return nullptr;
        ListNode* node = new ListNode(0);
        ListNode* p = node;
        ListNode* temp;
        while (pHead1 && pHead2) {
            if (pHead1->val <= pHead2->val) {
                temp = pHead1->next;
                p->next = pHead1;
                pHead1->next = nullptr;
                pHead1 = temp;
            } else {
                temp = pHead2->next;
                p->next = pHead2;
                pHead2->next = nullptr;
                pHead2 = temp;
            }
            p = p->next;
        }
        if (pHead1) {
            p->next = pHead1;
        }
        if (pHead2) {
            p->next = pHead2;
        }
        return node->next;
    }
};