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) {
        if (!pHead1) return pHead2;
        if (!pHead2) return pHead1;

        ListNode* dummy = new ListNode(0); // 创建一个虚拟头节点
        ListNode* current = dummy;

        while (pHead1 && pHead2) {
            if (pHead1->val < pHead2->val) {
                current->next = pHead1;
                pHead1 = pHead1->next;
            } else {
                current->next = pHead2;
                pHead2 = pHead2->next;
            }
            current = current->next;
        }

        if (pHead1) {
            current->next = pHead1;
        } else {
            current->next = pHead2;
        }

        ListNode* mergedList = dummy->next;
        delete dummy; // 释放虚拟头节点

        return mergedList;
    }
};