单词分类排序-基于二维链表

问题来源：http://topic.youkuaiyun.com/u/20110702/19/572f6747-20cb-45ac-8efc-b694820ded26.html?seed=158117872&r=74178770#r_74178770

每一行存储个数相同的字符串，行与行之间根据单词长度从大到小。

存储结构是个二维链表。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define  SIZE 100

struct RowNode 
{
	char str[SIZE];
	RowNode* next;
};


struct ORNode
{
	int strLen;
	RowNode* rowHead;
	ORNode* next;
};

void InitRowNode(RowNode* & node)
{
	memset(node->str, 0, SIZE);
	node->next = NULL;
}

void InitORNode(ORNode* & node)
{
	node->strLen = 0;
	node->rowHead = NULL;
	node->next = NULL;
}

/*
** 插入新单词
 */
void InsertWord(ORNode* & head, char* word)
{
	// 查找列
	ORNode* pLine = NULL;
	ORNode* preLine = NULL;
	ORNode* newLine = NULL;

	RowNode* pRow = NULL;
	RowNode* preNode = NULL;

	RowNode* pNode = (RowNode*)malloc( sizeof(RowNode) );
	InitRowNode(pNode);
	strcpy(pNode->str, word);

	for (pLine = head; pLine &&  (strlen(word)>pLine->strLen) ; preLine = pLine, pLine = pLine->next)
		;

	// pLine ==NULL时候：head为NULL 或者 word长度小于所有行的长度
	if ( pLine==NULL || strlen(word)<pLine->strLen )		// 新建一行插入
	{
		newLine = (ORNode*)malloc( sizeof(ORNode) );
		InitORNode(newLine);

		newLine->strLen = strlen(word);
		newLine->rowHead = pNode;

		if (preLine==NULL)			// 直接插在头结点位置
		{
			newLine->next = head;
			head = newLine;
		}
		else			// 插入preLine后一行
		{
			newLine->next = preLine->next;
			preLine->next = newLine;
		}
	}
	else								// 插入pLine指向的行
	{
		// pLine->rowHead为当前行
		// 把word按照字典顺序插入该行
		for (pRow = pLine->rowHead; pRow; preNode = pRow, pRow = pRow->next)
		{
			if ( strcmp(word, pRow->str)<0 )
				break;
		}

		if (preNode==NULL)		// 插入头部
		{
			pNode->next = pLine->rowHead;
			pLine->rowHead = pNode;
		}
		else					// 插入preNode的后面
		{
			pNode->next = preNode->next;
			preNode->next = pNode;
		}
	}
}

/*
** 按行打印单词
 */
void PrintWordsByLine(ORNode* head)
{
	ORNode* pIter = head;
	RowNode* pRowIter = NULL;

	while (pIter)
	{
		printf( "word of %d letter(s) :\n", pIter->strLen);

		pRowIter = pIter->rowHead;

		while (pRowIter)
		{
			printf("%s\n", pRowIter->str);
			pRowIter = pRowIter->next;
		}

		pIter = pIter->next;
	}
}

/*
** 销毁十字链表
 */
void DestroyOrthoList(ORNode* &head)
{
	ORNode* pIter = head;
	ORNode* pWork = NULL;

	RowNode* pRowIter = NULL;
	RowNode* pRowWork = NULL;

	while (pIter)
	{
		// 销毁行
		pRowIter = pIter->rowHead;

		while (pRowIter)
		{
			pRowWork = pRowIter;
			pRowIter = pRowIter->next;
			free(pRowWork);
			pRowWork = NULL;
		}

		// 销毁该结点
		pWork = pIter;
		pIter = pIter->next;
		free(pWork);
		pWork = NULL;
	}
}

int main()
{
	FILE* fp = fopen("input.txt", "r+");

	if (fp==NULL)
		exit(0);

	ORNode* head = NULL;

	int count=0;
	char str[SIZE];
	memset(str, 0, SIZE);

	while ( fscanf(fp, "%s", str)==1 )
	{
		InsertWord(head, str);

		++count;
		memset(str, 0, SIZE);
	}

	PrintWordsByLine(head);
	DestroyOrthoList(head);
	
	fclose(fp);

	system("PAUSE");
	return 0;
}

测试文件内容 input.txt：

void PrintWordsByLine(ORNode* head)
{
	ORNode* pIter = head;
	RowNode* pRowIter = NULL;

	while (pIter)
	{
		printf( "word of %d letter(s) :\n", pIter->strLen);

		pRowIter = pIter->rowHead;

		while (pRowIter)
		{
			printf("%s\n", pRowIter->str);
			pRowIter = pRowIter->next;
		}

		pIter = pIter->next;
	}
}

统计PrintWordsByLine函数这段代码，输出结果：

word of 1 letter(s) :
        =
        =
        =
        =
        =
        {
        {
        {
        }
        }
        }
word of 2 letter(s) :
        %d
        of
word of 4 letter(s) :
        void
word of 5 letter(s) :
        "word
        :\n",
        NULL;
        head)
        head;
        pIter
        pIter
        while
        while
word of 7 letter(s) :
        (pIter)
        ORNode*
        printf(
word of 8 letter(s) :
        RowNode*
        pRowIter
        pRowIter
        pRowIter
word of 9 letter(s) :
        letter(s)
word of 10 letter(s) :
        (pRowIter)
word of 12 letter(s) :
        pIter->next;
word of 14 letter(s) :
        printf("%s\n",
word of 15 letter(s) :
        pIter->rowHead;
        pIter->strLen);
        pRowIter->next;
        pRowIter->str);
word of 24 letter(s) :
        PrintWordsByLine(ORNode*
请按任意键继续. . .

示例存储结构图：

同时 参见之前一篇blog 基于C++的十字链表实现：http://blog.youkuaiyun.com/dizuo/article/details/6361077