字符串匹配_1

/*
問題描述：字符串匹配，暴力枚舉法和Rabin-Karp
來源：網易算法課
日期：2017-10-26
*/
#include <iostream>
#include <string>
using namespace std;

//暴力枚舉法
//target 目標字符串 str 待查找字符串 
int match_1(string target, string str)
{
	int i, j;
	for (i = 0; i < str.length() - target.length(); i++)
	{
		for (j = 0; j < target.length(); j++)
		{
			if (target[j] == str[i + j] && j == target.length() - 1)
				return i;
			else if (target[j] != str[i + j])
				break;
		}
	}
	return -1;
}

//Rabin-Karp
//target 目標字符串 str 待查找字符串 
//個人認爲有在字符串的組成上，Rabin-Karp要求字符串中元素要求是同一類型，有一定的侷限性
int match_2(string target, string str)
{
	//target預處理
	int i, j, targetValue = 0, strValue = 0;
	int h = 1;//最高位倍數
	for (i = 0; i < target.length() - 1; i++)
	{
		h = h * 26;
		h = h % 13;
	}


	for (i = 0; i < target.length(); i++)
	{
		targetValue = (26 * targetValue + (target[i] - 'a')) % 13;// %13防止整數過大越界
		strValue = (26 * strValue + (str[i] - 'a')) % 13;
	}

	for (i = 0; i < str.length() - target.length() + 1; i++)
	{
		if (targetValue == strValue)
		{
			for (j = 0; j < target.length(); j++)
			{
				if (j == target.length() - 1 && target[j] == str[i + j])
				{
					return i;
				}
				else if (target[j] != str[i + j])
				{
					break;
				}

			}
		}

		strValue = (26 * (strValue - (str[i] - 'a')*h) + str[i + target.length()] - 'a') % 13;
		if (strValue < 0)
		{
			strValue += 13;
		}
	}
	return -1;
}
/*
void main()
{
	string target = "aaabb";
	string str = "aaaccaaabbbaa";
	int index = match_2(target, str);
	string result = str.substr(index, target.length());
	cout << result << endl;

}
*/