每日一题(83) - 计算字符串的相似度

题目来自编程之美

题目

思路：其实就是求两个字符串的编辑距离，直接给出DP的思路

代码

#include <iostream>
#include <assert.h>
#include <string>
using namespace std;

/*
f[i][j]:表示字符串A[1~i]转换成B[1~j]的最小编辑距离
f[i][j] = min{f[i - 1][j - 1] + value,f[i][j - 1] + 1,f[i - 1][j] + 1};其中A[i] = B[j]，则value = 0，否则value = 1
初始化:f[0][j] = j;f[i][0] = i
结果：f[nLenA - 1][nLenB - 1]
*/
int f[50][50];

int Min(int x,int y,int z)
{
	if (x < y)
	{
		if (x < z)
		{
			return x;
		}
		else
		{
			return z;
		}
	}
	else
	{
		if (y < z)
		{
			return y;
		}
		else
		{
			return z;
		}
	}

}
int EditDistance(char* pStrA,char* pStrB)
{
	assert(pStrA && pStrB);
	int nLenA = strlen(pStrA);
	int nLenB = strlen(pStrB);
	int nValue = 0;
	//初始化
	for (int i = 0;i < nLenA;i++)
	{
		f[i][0] = i;
	}
	for (int j = 0;j < nLenB;j++)
	{
		f[0][j] = j;
	}
	
	for (int i = 1;i < nLenA;i++)
	{
		for (int j = 1;j < nLenB;j++)
		{
			nValue = 1;
			if (pStrA[i] == pStrB[j])
			{
				nValue = 0;
			}
			f[i][j] = Min(f[i - 1][j - 1] + nValue,f[i][j - 1] + 1,f[i - 1][j] + 1);
		}
	}
	return f[nLenA - 1][nLenB - 1];
}

int main()
{
	char strA[50];
	char strB[50];
	string str;
	//输入字符串A
	cin>>str;
	strA[0] = ' ';
	for (int i = 0;i < str.size();i++)
	{
		strA[i + 1] = str[i];
	}
	strA[str.size() + 1] = 0;
	//输入字符串A
	cin>>str;
	strB[0] = ' ';
	for (int i = 0;i < str.size();i++)
	{
		strB[i + 1] = str[i];
	}
	strB[str.size() + 1] = 0;

	int nSimNum = EditDistance(strA,strB);
	cout<< 1.0 / (nSimNum + 1)<<endl;
	system("pause");
	return 1;
}