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) {
        ListNode* res = (ListNode*)malloc(sizeof(ListNode));
        res->next = nullptr;
        ListNode* p=res;
        while (pHead1 != nullptr && pHead2 != nullptr) {
            if (pHead1->val < pHead2->val) {
                p->next = pHead1;
                pHead1 = pHead1->next;
            } else {
                p->next = pHead2;
                pHead2 = pHead2->next;
            }
            p=p->next;
        }
        while (pHead1 != nullptr) {
            p->next = pHead1;
            pHead1 = pHead1->next;
            p=p->next;
        }
        while (pHead2 != nullptr) {
            p->next = pHead2;
            pHead2 = pHead2->next;
            p=p->next;
        }
        return res->next;
    };
};