Distinct Subsequences

标签：动态规划   leetcode   distinct subsequence   

题目：

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.

这个题目可以用动态规划来解，既然是动态规划，所以关键是要找到递推关系。

我们设S(i)表示第i个字符之前的S的子串（不包含第i个字符），同理设T(j)表示第j个字符之前的T的子串。

设C(i,j)表示S(i)中有T(j)的个数。

于是就有递推关系：

if (当第S(i)的最后一个字符与T(j)的最后一个字符不等时)  C(i,j)=C(i-1,j)+C(i-1,i-1);

else C(i,j)=C(i-1,j) 

在编写程序时自然可以利用一个二维数组table[i][j]来表示C(i,j)的值。

java code:

public class DistinctSubsequences {
	public static int numDistincts(String S, String T) {
		int[][] table = new int[S.length() + 1][T.length() + 1];
		for (int i = 0; i <= S.length(); i++) {
			table[i][0] = 1;// 初始化S到T的空子串为1
		}
		for (int i = 1; i <= S.length(); i++) {
			for (int j = 1; j <= T.length(); j++) {
				if (S.charAt(i - 1) == T.charAt(j - 1)) {
					table[i][j] = table[i - 1][j - 1] + table[i - 1][j];
				} else {
					table[i][j] = table[i - 1][j];
				}
			}
		}
		return table[S.length()][T.length()];
	}

	public static void main(String[] args) {
		// String S = "b", T = "b";
		// String S = "abc", T = "";
		String S = "rabbbit", T = "rabbit";
		System.out.println(numDistincts(S, T));
	}
}

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Distinct Subsequences

标签：动态规划   leetcode   distinct subsequence   

原文地址：http://blog.csdn.net/louxuez/article/details/24911841