leetcode 21. 合并两个有序链表

我的解题

#include <bits/stdc++.h>

using namespace std;

/**
 * Definition for singly-linked list.
 */
struct ListNode
{
    int val;
    ListNode *next;
    ListNode() :
                    val(0), next(nullptr)
    {
    }
    ListNode(int x) :
                    val(x), next(nullptr)
    {
    }
    ListNode(int x, ListNode *next) :
                    val(x), next(next)
    {
    }
};

class Solution
{
public:
    ListNode* mergeTwoLists(ListNode *a, ListNode *b)
    {
        if ((!a) || (!b))
            return a ? a : b;
        /**
         * head是"dummy node for create linked list"
         * 需要注意的是它是automatic variable，使用它为了便于创建新的linked list
         * 最终返回的 head.next
         */
        ListNode head, *tail = &head;
        ListNode *aPtr = a, *bPtr = b;
        while (aPtr && bPtr)
        {
            if (aPtr->val < bPtr->val)
            {
                tail->next = aPtr;
                aPtr = aPtr->next; // 迭代
            }
            else
            {
                tail->next = bPtr;
                bPtr = bPtr->next; // 迭代
            }
            tail = tail->next;
        }
        tail->next = (aPtr ? aPtr : bPtr); // 将剩余部分也添加到linked list中
        return head.next;
    }
};

ostream& operator<<(ostream &stream, ListNode *Node)
{
    while (Node)
    {
        stream << Node->val << ",";
        Node = Node->next;
    }
    return stream;
}

int main()
{
    ListNode *N1 = new ListNode { 1 };
    N1->next = new ListNode { 3 };
    N1->next->next = new ListNode { 5 };
    N1->next->next->next = new ListNode { 7 };

    ListNode *N2 = new ListNode { 2 };
    N2->next = new ListNode { 4 };
    N2->next->next = new ListNode { 6 };
    N2->next->next->next = new ListNode { 8 };

    Solution s;
    cout << s.mergeTwoLists(N1, N2) << endl;
}