标签：style color io os ar java for strong sp
Given a string S and a string T, count the number of distinct subsequences of T in S.
A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie,
"ACE" is a subsequence of "ABCDE" while "AEC" is not).
Here is an example:
S = "rabbbit", T = "rabbit"
Return 3.
public class Solution {
int[][] matrix;
boolean []equals;
String S;
String T;
int lenS;
int lenT;
private int calDistinct(int i, int j)
{
if (matrix[i][j] != -1)
{
return matrix[i][j];
}
if (lenS - i == lenT - j)
{
matrix[i][j] = equals[j]? 1 : 0;
return matrix[i][j];
}
if (lenS - i < lenT - j)
{
matrix[i][j] = 0;
return matrix[i][j];
}
matrix[i][j] = 0;
if (S.charAt(i) == T.charAt(j))
{
if (j + 1 == lenT)
{
matrix[i][j] += 1;
}
else
{
matrix[i][j] += calDistinct(i + 1, j + 1);
}
}
for(int p=i+1;p<lenS;p++)
{
if(S.charAt(p)==T.charAt(j))
{
matrix[i][j] += calDistinct(p, j);
break;
}
}
return matrix[i][j];
}
public int numDistinct(String S, String T)
{
if (S == null || T == null || S.length() < T.length())
{
return 0;
}
if (T.length() == 0)
{
return 1;
}
if (S.length() == 0)
{
return 0;
}
this.S = S;
this.T = T;
this.lenS = S.length();
this.lenT = T.length();
matrix = new int[lenS][lenT];
for (int i = 0; i < lenS; i++ )
{
for (int j = 0; j < lenT; j++ )
{
matrix[i][j] = -1;
}
}
equals=new boolean[lenT];
equals[lenT-1]=S.charAt(lenS-1)==T.charAt(lenT-1);
int p;
for(p=lenT-2;p>=0;p--)
{
equals[p]=equals[p+1]&&S.charAt(lenS-(lenT-p))==T.charAt(p);
}
return calDistinct(0, 0);
}
}标签：style color io os ar java for strong sp
原文地址：http://blog.csdn.net/jiewuyou/article/details/40107721
