Description
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.
Solution
跟Edit distance思路很像的DP问题。用dp[i][j]代表s.substring(0, i)中t.substring(0, j)的子序列个数。
class Solution {
public int numDistinct(String s, String t) {
int m = s.length();
int n = t.length();
if (m < n) {
return 0;
}
int[][] dp = new int[m + 1][n + 1];
for (int i = 0; i <= m; ++i) {
for (int j = 0; j <= n; ++j) {
if (j == 0) {
dp[i][j] = 1;
} else if (i < j) {
dp[i][j] = 0;
} else {
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[m][n];
}
}
