Distinct Subsequences (Java)

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.

字符串的匹配以及子串問題，要首先想到DP。

這道題考察的是從S刪除若干字符後與T相等的方法有多少種，用dp[i][j]表示從0到i，從0到j的變換方式（刪除方法）有多少種。

參考http://www.cnblogs.com/TenosDoIt/p/3440022.html

Source

public class Solution {
    public int numDistinct(String S, String T) {
    	int len1 = S.length(), len2 = T.length();
    	if(len1 == 0) return 0;
    	if(len2 == 0) return 1;    //T爲空串，變化方式爲1，空串是任何字符串的子串
    	
    	int[][] dp = new int[len1 + 1][len2 + 1];
    	
    	for(int i = 0; i <= len1; i++){
    		dp[i][0] = 1;
    	}
    	
    	for(int i = 1; i <= len1; i++){
    		for(int j = 1; j <= len2; 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];  //至少等於dp[i-1][j]
    			//這道題考察的是S刪除某些位置得T的方法有多少種，所以遍歷時肯定會出現當前子串不相等的情況，比如
    			//S=rabbb T=rabbi的這種情況 此時dp[i][j] = dp[i-1][j] 刪除方法和上一步i-1時相同
    		}
    	}
    	return dp[len1][len2];
    }
}