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) {}
 * };
 */
#include <cstddef>
#include <cstdlib>
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 = (ListNode*)malloc(sizeof(ListNode));
        ListNode* end = head;
        // 带头结点的单链表是真的好操作
        while (pHead1 != NULL && pHead2 != NULL) {
            if (pHead1 -> val < pHead2 ->val) {
                end -> next = pHead1;
                end = pHead1;
                pHead1 = pHead1 -> next;
            } else {
                end -> next = pHead2;
                end = pHead2;
                pHead2 = pHead2 -> next;
            }
        }
        end -> next = pHead1 ? pHead1 : pHead2;
        return head -> next;
    }
};