lintcode-medium-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).

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

 

动态规划，用一个int二维数组记录：字符串T中前j个字符（0-j-1）在字符串S中前i个字符（0-i-1）中不同的子序列的个数

状态转移：

对任何一种情况，T中前j个字符在S中前i个字符出现的次数，至少等于T中前j个字符在S中前i-1个字符的次数，这里包含一个特殊情况，如果j == i，这个值为0

如果T的第j - 1个字符和S的i - 1个字符不同，所以S中再增加一个字符不会改变结果，则dp[i][j] = dp[i - 1][j]

如果相同，则dp[i][j] 在dp[i - 1][j]的基础上，还要再包括T中前j-1个字符在S中前i-1个字符中出现的次数（即先把两个字符串的最后一个字符匹配到一起，再考虑之前的情况），dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1]

public class Solution {
    /**
     * @param S, T: Two string.
     * @return: Count the number of distinct subsequences
     */
    public int numDistinct(String S, String T) {
        // write your code here
        
        if(S == null || S.length() == 0)
            return 0;
        
        if(T == null || T.length() == 0)
            return 1;
        
        int m = S.length();
        int n = T.length();
        
        if(n > m)
            return 0;
        
        int[][] dp = new int[m + 1][n + 1];
        
        dp[0][0] = 1;
        for(int i = 0; i <= m; i++)
            dp[i][0] = 1;
        
        for(int j = 1; j <= n; j++){
            char T_end = T.charAt(j - 1);
            
            for(int i = j; i <= m; i++){
                char S_end = S.charAt(i - 1);
                
                dp[i][j] = dp[i - 1][j];
                if(S_end == T_end)
                    dp[i][j] += dp[i - 1][j - 1];
            }
        }
        
        return dp[m][n];
    }
}