Given a string S and a string T, count the number of distinct subsequences of S which equals T.
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 int numDistinct(String s, String t) {
int slen = s.length(), tlen = t.length();
int[][] dp = new int[slen + 1][tlen + 1];
dp[0][0] = 1;
for (int i = 1; i <= slen; i++) {
dp[i][0] = 1;
}
for (int i = 1; i <= slen; i++) {
for (int j = 1; j <= tlen; j++) {
dp[i][j] = dp[i - 1][j];
if (s.charAt(i - 1) == t.charAt(j - 1)) dp[i][j] += dp[i - 1][j - 1];
}
}
return dp[slen][tlen];
}
本文介绍了一种算法,用于计算一个字符串作为另一个字符串子序列出现的次数。通过使用动态规划的方法,该算法能有效地解决这一问题。示例中给出了具体的实现代码。
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