剑指offer 48： 最长不含重复字符的子字符串

使用循环实现。其中创建了一个长度为26的辅助数组，存放各字符在字符串中最后出现的位置。

/**
 *Copyright @ 2019 Zhang Peng. All Right Reserved.
 *Filename:
 *Author: Zhang Peng
 *Date:
 *Version:
 *Description:剑指offer<第二版> 面试题48 最长不含重复字符的子字符串
**/

#include
using namespace std;

int findLongStr(char * str)
{
	
	int s = strlen(str);
	if (s == 0)
		return 0;
	int record[26];
	for (int i = 0; i < 26; i++)
		record[i] = -1;
	int start = 0;
	record[str[0] - 'a'] = 0;
	int maxlen = 1;
	for (int i = 1; i < s; i++)
	{
		int findindex = str[i] - 'a';
		int len = 0;
		if (record[findindex] == -1)
			len = i - start + 1;
		else
		{	
			if (record[findindex]<start)
				len = i - start + 1;
			else
			{
				len = i - record[findindex];
				start = record[findindex]+1;
			}		
		}
		if (len>=maxlen)
		{
			cout << "length " << len << " str:" << endl;
			maxlen = len;
			for (int k = start; k < (start+maxlen); k++)
				cout << str[k];
			cout << endl;
		}	
		record[findindex] = i;   //记录最新的字符所处位置
	}
	return maxlen;
}


int main()
{
	char * testcase = "arabcaacfr";
	int num = findLongStr(testcase);

	cout << "max num: " << num << endl;

    system("pause");
    return 0;
}

运行结果如下：