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 <iostream>
#include <queue>
#include <stack>
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

        //解题思路 比较大小之后，立即将该结点作为新的结点插入操新的队列（使用尾插法）
        //空间复杂度1 时间复杂度n

        //返回的链表
        auto newList = new ListNode(0);
        ListNode* rear = newList;

        //先逐个比较两个队列中的元素
        while (pHead1  && pHead2 ) {
            if (pHead1->val <= pHead2->val) {
                auto node = new ListNode(pHead1->val);
                rear->next= node;
                rear = node;
                pHead1 = pHead1->next;
            } else {
                auto node = new ListNode(pHead2->val);
               rear->next= node;
                rear = node;
                pHead2 = pHead2->next;
            }
        }
        //若比较完有剩余，则将链表中剩余元素挨个添加进链表
        while (pHead1  || pHead2 ) {
            if (pHead1 != nullptr) {
                auto node = new ListNode(pHead1->val);
                rear->next= node;
                rear = node;
                pHead1 = pHead1->next;
            }
            if (pHead2 != nullptr) {
                auto node = new ListNode(pHead2->val);
                rear->next= node;
                rear = node;
                pHead2 = pHead2->next;
            }
        }

        return newList->next;
    }
};