Distinct Subsequences

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).

Example

Given S = "rabbbit", T = "rabbit", return 3.

递归：

if (S.charAt(indexS) == T.charAt(indexT)) {
　　return numDistinct(S, indexS + 1, T, indexT + 1) + numDistinct(S, indexS + 1, T, indexT);
} else {
　　return numDistinct(S, indexS + 1, T, indexT);
}

DP:

if (S.charAt(i - 1) == T.charAt(j - 1)) {
　　dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] (不考虑S中第i + 1个字符);
} else {
　　dp[i][j] = dp[i - 1][j]; 
}

 

public class Solution {
    /**
     * @param S, T: Two string.
     * @return: Count the number of distinct subsequences
     */
     
    /*  over time
    public int numDistinct(String S, String T) {
        if (T == null) return 1;
        if (S == null) return 0;
        
        return numDistinct(S, 0, T, 0);
    }
    
    private int numDistinct(String S, int indexS, String T, int indexT) {
        if (indexT >= T.length()) return 1;
        if (indexS >= S.length()) return 0;
        
        if (S.charAt(indexS) == T.charAt(indexT)) {
            return numDistinct(S, indexS + 1, T, indexT + 1) + numDistinct(S, indexS + 1, T, indexT);
        } else {
            return numDistinct(S, indexS + 1, T, indexT);
        }
    }
    */
    
    public int numDistinct(String S, String T) {
        if (T == null) return 1;
        if (S == null) return 0;
        
        int[][] dp = new int[S.length() + 1][T.length() + 1];
        
        for (int i = 1; i < dp[0].length; i++) {
            dp[0][i] = 0;
        }
        
        for (int i = 0; i < dp.length; i++) {
            dp[i][0] = 1;
        }
        
        for (int i = 1; i < dp.length; i++) {
            for (int j = 1; j < dp[0].length; j++) {
                if (S.charAt(i - 1) == T.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        return dp[dp.length - 1][dp[0].length - 1];
    }
}