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
        ListNode* head = new ListNode(0);
        ListNode* p = head;
        while (phead1 != nullptr && phead2 != nullptr) {
            if (phead1->val > phead2->val) {
                p->next = phead2;
                phead2 = phead2->next;
            } else {
                p->next = phead1;
                phead1 = phead1->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;
        }
        ListNode* re=head->next;
        delete(head);
        return re;
    }
};