LCS最长子串问题

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方法一：动态规划 效率O（mn）（mn是分别是两个字符串的长度）

#include<iostream>
using namespace std;

int c[100][100];//全局变量自动初始化为0
inline int max(int a, int b)
{
	return (a > b ? a : b);
}

int LCS(const char *X,const char *Y)
{
	if (NULL == X || NULL == Y)
		return 0;
	int xlen = strlen(X);
	int ylen = strlen(Y);
	for (int i = 1; i <= xlen; i++)
	{
		for (int j = 1; j <= ylen; j++)
		{
			if (X[i] == Y[j])
				c[i][j] = c[i - 1][j - 1] + 1;
			else
				c[i][j] = max(c[i][j - 1], c[i - 1][j]);
		}
	}
	return c[xlen][ylen];
}
int main()
{
	char A[8] = "ABCBDAB";
	char B[8] = "BDCABAB";
	cout << LCS(A, B) << endl;
	system("pause");
	return 0;
}

#define INF 9999999
int c[100][100];
int LCS_Memo(const char* X, const char* Y,int i,int j)
{
	if (c[i][j] < INF)
		return c[i][j];
	if (0 == i || 0 == j)
		c[i][j] = 0;
	else if (X[i - 1] == Y[j - 1])
		c[i][j] = LCS_Memo(X,Y,i - 1, j - 1) + 1;
	else
	{
		int p = LCS_Memo(X, Y, i - 1, j);
		int q = LCS_Memo(X, Y, i, j - 1);
		if (p >q)
			c[i][j] = p;
		else
		{
			c[i][j] = q;
		}
	}
	return c[i][j];
}
int LCS(const char* X ,const char* Y)
{
	if (NULL == X || NULL == Y)
		return 0;
	int xlen = strlen(X);
	int ylen = strlen(Y);
	memset(c, INF, sizeof(c));//注意是将每个字节赋值为INF，不是一个int
	
	return LCS_Memo(X, Y, xlen, ylen);
}
int main()
{
	char A[8] = "ABCBDAB";
	char B[8] = "BDCABAB";
	cout << LCS(A, B) << endl;
	system("pause");
	return 0;
}

LCS最长子串问题

标签：

原文地址：http://blog.csdn.net/iamxiaoguizi/article/details/51201156