获取一个字符串的最长无重复字符子串

//规则很简单，获取一个字符串的最长子串，要求这个子串中没有重复字符。字符穿中的字符均为ASCII字符
//下面的代码获取了所有最长无重复子串，即如果有多个满足条件的子串长度都相同，则全部输出
//如果只需要输出第一条最长子串，注释掉nTempLen == selData.m_Len判定分支的代码即可
//分析过程随后补上


#include "stdafx.h"
#include <string>
#include <vector>
using namespace std;

#ifndef byte
#define byte unsigned char
#endif

class SubStrData
{
public:
	SubStrData(int beginPos, int len);
	~SubStrData() {}

public:
	int m_BeginPos;
	int m_Len;
};

SubStrData::SubStrData(int beginPos = 0, int len = 0)
{
	m_BeginPos = beginPos;
	m_Len = len;
}

void GetMaxSubString(const char * strOrig, vector<SubStrData> & datas);

int _tmain(int argc, _TCHAR* argv[])
{
	char buffIn[1024];
	char buffOut[1024];

	char *pch;
	while ((pch = gets_s(buffIn, 1024)) != NULL)
	{
		vector<SubStrData> datas;
		GetMaxSubString(buffIn, datas);
		printf("Sub String Count = %d:\n", datas.size());
		for (int i=0; i<datas.size(); i++)
		{
			strncpy_s(buffOut, buffIn + datas[i].m_BeginPos, datas[i].m_Len);
			printf("Max SubString pos=%d, len=%d\n", datas[i].m_BeginPos, datas[i].m_Len);
			printf("%s\n", buffOut);
		}
	}

	return 0;
}

void GetMaxSubString(const char * strOrig, vector<SubStrData> & datas)
{
	int len = strlen(strOrig);
	int map[256]; 
	for (int i=0; i<256; i++)
	{
		map[i] = -1;
	}
	
	SubStrData selData;
	int nMaxLen = 0;
	int nCheckingBegin = 0;
	for (int i=0; i<len; i++)
	{
		if (map[(byte)strOrig[i]] == -1 || map[(byte)strOrig[i]] < nCheckingBegin)
		{
			map[(byte)strOrig[i]] = i;
		}
		else
		{
			int nTempLen = i - nCheckingBegin;
			if (nTempLen > selData.m_Len)
			{
				selData.m_Len = nTempLen;
				selData.m_BeginPos = nCheckingBegin;

				datas.clear();
				SubStrData dataItem = selData;
				datas.push_back(dataItem);
			}
			else if (nTempLen == selData.m_Len)
			{
				SubStrData dataItem(nCheckingBegin, nTempLen);
				datas.push_back(dataItem);
			}
			nCheckingBegin = map[(byte)strOrig[i]] + 1;
			map[(byte)strOrig[i]] = i;
		}
	}

	int nTempLen = len - nCheckingBegin;
	if (nTempLen > selData.m_Len)
	{
		selData.m_Len = nTempLen;
		selData.m_BeginPos = nCheckingBegin;

		datas.clear();
		SubStrData dataItem = selData;
		datas.push_back(dataItem);
	}
	else if (nTempLen == selData.m_Len)
	{
		SubStrData dataItem(nCheckingBegin, nTempLen);
		datas.push_back(dataItem);
	}

}