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
        if (pHead1 == nullptr && pHead2 == nullptr) {
            return nullptr;
        } else if (pHead1 == nullptr) {
            return pHead2;
        } else if (pHead2 == nullptr) {
            return pHead1;
        }

        ListNode* p = nullptr;
        ListNode* q = nullptr;
        if (pHead1->val < pHead2->val) {
            p = pHead1;
            q = pHead2;
        } else {
            p = pHead2;
            q = pHead1;
        }
        ListNode* head = p;
        ListNode* r;

        while (!(p->next == nullptr && q->next == nullptr)) {
            if (q == nullptr) break;
            if (p->val <= q->val && (p->next->val > q->val || p->next == nullptr)) {
                r = q->next;
                q->next = p->next;
                p->next = q;
                p = q;
                q = r;
            } else if (p->val <= q->val && p->next->val <= q->val) {
                p = p->next;
            } else  {
                q = q->next;
            }
        }
        if (p->next == nullptr) {
            p->next = q;
        }
        return head;
    }
};