每日一题(92) - 归并排序

最新推荐文章于 2024-03-21 18:55:00 发布

原创最新推荐文章于 2024-03-21 18:55:00 发布 · 1.1k 阅读

0 ·

CC 4.0 BY-SA版权

面试题专栏收录该内容

108 篇文章

订阅专栏

本文深入讲解了归并排序算法的两种实现方式：基于数组和基于链表。详细介绍了每种方式的具体实现步骤、代码示例及注意事项。对于理解归并排序的时间复杂度和空间复杂度具有较高的参考价值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目来自网络

题目(1)：基于数组的归并排序

题目(2)：基于链表的归并排序

----------

题目(1)：基于数组的归并排序

思路：借助分治的思想，总是先对待排序进行分割，之后再对两个子序列进行合并，生成一个有序序列。

时间复杂度为O(nLogn),空间复杂度为O（n）

代码：

void Merge(int nArr[],int nStart,int nMid,int nEnd)
{
	int nCurLeft = nStart;
	int nLeftEnd = nMid;
	int nCurRight = nMid + 1;
	int nRightEnd = nEnd;

	int nCur = 0;//易错点
	int* nArrMerge = new int[nEnd - nStart + 1];
	while(nCurLeft <= nLeftEnd && nCurRight <= nRightEnd)
	{
		if (nArr[nCurLeft] > nArr[nCurRight])
		{
			nArrMerge[nCur++] = nArr[nCurRight++];
		}
		else
		{
			nArrMerge[nCur++] = nArr[nCurLeft++];
		}
	}
	while (nCurLeft <= nLeftEnd)
	{
		nArrMerge[nCur++] = nArr[nCurLeft++]; 
	}
	while(nCurRight <= nRightEnd)
	{
		nArrMerge[nCur++] = nArr[nCurRight++];
	}
	for (int i = 0;i < nCur;i++)//易忘点
	{
		nArr[i + nStart] = nArrMerge[i];
	}
	delete[] nArrMerge;
}

void MergeSort(int nArr[],int nStart,int nEnd)
{
	if (nStart >= nEnd)
	{
		return;
	}
	int nMid = (nStart + nEnd) >> 1;
	MergeSort(nArr,nStart,nMid);
	MergeSort(nArr,nMid + 1,nEnd);
	Merge(nArr,nStart,nMid,nEnd);
}

void MergeSort(int nArr[],int nLen)
{
	assert(nArr != NULL && nLen > 0);
	MergeSort(nArr,0,nLen - 1);
}

注意：下面代码的辅助空间是每次归并都申请一次，当然也可以只申请一次。这里代码就不表了。

题目(2)：基于链表的归并排序

代码

#include <iostream>
using namespace std;

struct ListNode  
{  
	int m_nData;  
	ListNode* m_pNext;  
}; 
ListNode* Merge(ListNode* pFirst,ListNode* pSec)
{
	ListNode* pNewHead = NULL;
	ListNode* pCur = NULL;
	while(pFirst && pSec)
	{
		if (pFirst->m_nData > pSec->m_nData)
		{
			if (pNewHead == NULL)
			{
				pNewHead = pSec;
				pCur = pSec;
			}
			else
			{
				pCur->m_pNext = pSec;
				pCur = pCur->m_pNext;
			}
			pSec = pSec->m_pNext;
		}
		else
		{
			if (pNewHead == NULL)
			{
				pNewHead = pFirst;
				pCur = pFirst;
			}
			else
			{
				pCur->m_pNext = pFirst;
				pCur = pCur->m_pNext;
			}
			pFirst = pFirst->m_pNext;
		}
	}
	if (pFirst == NULL && pSec != NULL)
	{
		pCur->m_pNext = pSec;
	}
	else
	{
		pCur->m_pNext = pFirst;
	}
	return pNewHead;
}

ListNode* MergeSort(ListNode* pHead)
{
	ListNode* pFirstHead = NULL;
	ListNode* pSecHead = NULL;

	if (pHead == NULL || pHead->m_pNext == NULL)
	{
		return pHead;
	}
	if (pHead->m_pNext->m_pNext == NULL)//只有俩元素
	{
		if (pHead->m_nData > pHead->m_pNext->m_nData)
		{
			ListNode* pSec = pHead->m_pNext;
			pHead->m_pNext = NULL;
			pSec->m_pNext = pHead;
			return pSec;
		}
		return pHead;		
	}
	//此时，链表肯定有三个以上的元素
	//分成两个子问题
	ListNode* pFast = pHead;
	ListNode* pSlow = pHead;
	while(pFast->m_pNext && pFast->m_pNext->m_pNext)
	{
		pSlow = pSlow->m_pNext;
		pFast = pFast->m_pNext->m_pNext;
	}
	pSecHead = pSlow->m_pNext;
	pSlow->m_pNext = NULL;
	pFirstHead = MergeSort(pHead);
	pSecHead = MergeSort(pSecHead);
	//子问题合并
	return Merge(pFirstHead,pSecHead);
}
void Print(ListNode* pHead)
{
	while(pHead != NULL)
	{
		cout<<pHead->m_nData<<" ";
		pHead = pHead->m_pNext;
	}
	cout<<endl;
}
void CreateList(ListNode** pHead,int nLen)//头指针使用指针的指针  
{  
	ListNode* pCur = NULL;  
	ListNode* pNewNode = NULL;  
	for (int i = 0;i < nLen;i++)  
	{  
		pNewNode = new ListNode;  
		cin>>pNewNode->m_nData;  
		pNewNode->m_pNext = NULL;  

		if (*pHead == NULL)  
		{  
			*pHead = pNewNode;  
			pCur = *pHead;  
		}  
		else  
		{  
			pCur ->m_pNext = pNewNode;  
			pCur = pNewNode;  
		}  
	}  
}  

int main()
{
	int nLen = 5;
	ListNode* pHead = NULL;
	CreateList(&pHead,nLen);
	pHead = MergeSort(pHead);
	Print(pHead);
	system("pause");
	return 1;
}