计算字符串的相似度---编辑距离

编辑距离：

又称Levenshtein距离（也叫做Edit Distance），是指两个字串之间，由一个转成另一个所需的最少编辑操作次数。许可的编辑操作包括将一个字符替换成另一个字符，插入一个字符，删除一个字符。

例如将kitten一字转成sitting：

sitten（k→s）

sittin（e→i）

sitting（→g）

利用递归来求解算法：

#include <iostream>

using namespace std;

int cacDistance(const char *lhs, int lBegin, int lEnd,
				const char *rhs, int rBegin, int rEnd)
{
	if (lBegin > lEnd)
	{
		if (rBegin > rEnd)
			return 0;
		else
			return rEnd - rBegin + 1;
	}
	if (rBegin > rEnd)
	{
		if (lBegin > lEnd)
			return 0;
		else
			return lEnd - lBegin + 1;
	}

	if (*(lhs + lBegin) == *(rhs + rBegin))
		return cacDistance(lhs, lBegin + 1, lEnd, rhs, rBegin + 1, rEnd);
	else
	{
		int t1 = cacDistance(lhs, lBegin + 1, lEnd, rhs, rBegin + 1, rEnd);
		int t2 = cacDistance(lhs, lBegin + 1, lEnd, rhs, rBegin, rEnd);
		int t3 = cacDistance(lhs, lBegin, lEnd, rhs, rBegin + 1, rEnd);
		return min(t1, min(t2, t3)) + 1;
	}
}

void main()
{
	const char *str1 = "kitten";
	const char *str2 = "sitting";
	int result = cacDistance(str1, 0, strlen(str1) - 1, str2, 0, strlen(str2) - 1);
	cout << "result: " << result << endl;
}

用动态规划来解此题，代码如下：

#include <iostream>

using namespace std;

int info[10][10];

void print(int size1, int size2)
{
	for (int i = 1; i <= size1; i++)
	{
		for (int j = 1; j <= size2; j++)
		{
			cout << info[i][j] << " ";
		}
		cout << endl;
	}
}

int cacDistance(const char *str1, const char *str2)
{
	if (str1 == NULL || str2 == NULL)
		return -1;

	int size1 = strlen(str1);
	int size2 = strlen(str2);
	for (int i = 0; i < size1; i++)
		info[i][size2] = 0;
	for (int j = 0; j < size2; j++)
		info[size1 - 1][j] = 0;

	for (int i = 1; i <= size1; i++)
	{
		for (int j = 1; j <= size2; j++)
		{
			if (str1[i - 1] == str2[j - 1])
				info[i][j] = info[i - 1][j - 1];
			else if (str1[i - 1] != str2[j - 1])
				info[i][j] = min(info[i - 1][j - 1], min(info[i - 1][j], info[i][j - 1])) + 1;
		}
	}

	print(size1, size2);
	return info[size1][size2];
}

void main()
{
	const char *str1 = "kitten";
	const char *str2 = "sitting";
	int result = cacDistance(str1, str2);
	cout << "result: " << result << endl;
}

递归的建表解法，也就是书上说的避免重复计算解法

#include <iostream>

using namespace std;

int info[10][10];
int INF = -1;

int cacDistance(const char *lhs, int lBegin, int lEnd,
				const char *rhs, int rBegin, int rEnd)
{
	if (lBegin == lEnd)
	{
		return info[lBegin + 1][rBegin + 1] + rEnd - rBegin;
	}
	else if (rBegin == rEnd)
	{
		return info[lBegin + 1][rBegin + 1] + lEnd - lBegin;
	}

	if (info[lBegin + 1][rBegin + 1] != -1)
	{
		return info[lBegin + 1][rBegin + 1];
	}

	if (*(lhs + lBegin) == *(rhs + rBegin))
		info[lBegin][rBegin] = cacDistance(lhs, lBegin + 1, lEnd, rhs, rBegin + 1, rEnd);
	else
	{
		int t1 = cacDistance(lhs, lBegin + 1, lEnd, rhs, rBegin + 1, rEnd);
		int t2 = cacDistance(lhs, lBegin + 1, lEnd, rhs, rBegin, rEnd);
		int t3 = cacDistance(lhs, lBegin, lEnd, rhs, rBegin + 1, rEnd);
		info[lBegin][rBegin] = min(t1, min(t2, t3)) + 1;
	}

	return info[lBegin][rBegin];
}

void main()
{
	const char *str1 = "kitten";
	const char *str2 = "sitting";
	int size1 = strlen(str1);
	int size2 = strlen(str2);
	for (int i = 0; i <= size1; i++)
		info[i][size2 - 1] = 0;
	for (int j = 0; j <= size2; j++)
		info[size1 - 1][j] = 0;

	for (int i = 1; i <= size1; i++)
	{
		for (int j = 1; j <= size2; j++)
		{
			info[i][j] = -1;
		}
	}
	int result = cacDistance(str1, 0, size1, str2, 0, size2);
	cout << "result: " << result << endl;
}