面试题25：合并两个排序的链表(C++/Python)

一、题目

    输入两个递增排序的链表，合并这两个链表并使新链表中的结点仍然是按照递增排序的。例如输入图3.11中的链表1和链表2，则合并之后的升序链表如链表3所示。

二、解法

2.1方法一：递归

#include <cstdio>
#include <iostream>

using namespace std;

struct ListNode
{
	int       m_nValue;
	ListNode* m_pNext;
};
ListNode* CreateListNode(int value)
{
	ListNode* pNode = new ListNode();
	pNode->m_nValue = value;
	pNode->m_pNext = nullptr;

	return pNode;
}

void ConnectListNodes(ListNode* pCurrent, ListNode* pNext)
{
	if (pCurrent == nullptr)
	{
		printf("Error to connect two nodes.\n");
		exit(1);
	}

	pCurrent->m_pNext = pNext;
}

ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
	if (pHead1 == nullptr)
		return pHead2;
	else if (pHead2 == nullptr)
		return pHead1;

	ListNode* pMergedHead = nullptr;

	if (pHead1->m_nValue < pHead2->m_nValue)
	{
		pMergedHead = pHead1;
		pMergedHead->m_pNext = Merge(pHead1->m_pNext, pHead2);
	}
	else
	{
		pMergedHead = pHead2;
		pMergedHead->m_pNext = Merge(pHead1, pHead2->m_pNext);
	}

	return pMergedHead;
}
int main()
{
	ListNode* pNode1 = CreateListNode(1);
	ListNode* pNode3 = CreateListNode(3);
	ListNode* pNode5 = CreateListNode(5);

	ConnectListNodes(pNode1, pNode3);
	ConnectListNodes(pNode3, pNode5);

	ListNode* pNode2 = CreateListNode(2);
	ListNode* pNode4 = CreateListNode(4);
	ListNode* pNode6 = CreateListNode(6);

	ConnectListNodes(pNode2, pNode4);
	ConnectListNodes(pNode4, pNode6);

	ListNode* pMergedHead = Merge( pNode1, pNode2);

	while (pMergedHead != nullptr)
	{
		cout << pMergedHead->m_nValue;
		pMergedHead = pMergedHead->m_pNext;
	}

	system("pause");
}

2.2 方法二：循环

    def Merge(self, pHead1, pHead2):
        # write code here
        if pHead1 == None:
            return pHead2
        if pHead2 == None:
            return pHead1
        pHead = ListNode(0)
        ptr = pHead
        while pHead1 != None and pHead2 != None:
            if pHead1.val <= pHead2.val or pHead2 == None:
                pTemp = pHead1
                pHead1 = pHead1.next if pHead1 != None else None
            else:
                pTemp = pHead2
                pHead2 = pHead2.next if pHead2 != None else None
            ptr.next = pTemp
            ptr = ptr.next
        if pHead1 == None:
            ptr.next = pHead2
        if pHead2 ==None:
            ptr.next = pHead1
        return pHead.next