Distinct Subsequences leetcode java

題目：

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.

 

題解：

 這道題首先引用我忘記在哪裡看到的一句話：

 “When you see string problem that is about subsequence or matching, dynamic programming method should come to your mind naturally. ”

 所以這種型別題可以多往DP思考思考。

 首先設定動態規劃陣列dp[i][j]，表示S串中從開始位置到第i位置與T串從開始位置到底j位置匹配的子序列的個數。

 如果S串為空，那麼dp[0][j]都是0；

 如果T串為空，那麼dp[i][j]都是1，因為空串為是任何字串的字串。

 可以發現規律，dp[i][j] 至少等於 dp[i][j-1]。

 當i=2，j=1時，S 為 ra，T為r，T肯定是S的子串；這時i=2，j=2時，S為ra，T為rs，T現在不是S的子串，當之前一次是子串所以現在計數為1.

 

同時，如果字串S[i-1]和T[j-1]（dp是從1開始計數，字串是從0開始計數）匹配的話，dp[i][j]還要加上dp[i-1][j-1]

 例如對於例子： S = "rabbbit", T = "rabbit"

 當i=2，j=1時，S 為 ra，T為r，T肯定是S的子串；當i=2，j=2時，S仍為ra，T為ra，這時T也是S的子串，所以子串數在dp[2][1]基礎上加dp[1][1]。

 

程式碼如下：

 1     public int numDistinct(String S, String T) {
 2         int[][] dp = new int[S.length() + 1][T.length() + 1];
 3         dp[0][0] = 1;//initial
 4         
 5         for(int j = 1; j <= T.length(); j++)//S is empty
 6             dp[0][j] = 0;
 7             
 8         for (int i = 1; i <= S.length(); i++)//T is empty
 9             dp[i][0] = 1;
10            
11         for (int i = 1; i <= S.length(); i++) {
12             for (int j = 1; j <= T.length(); j++) {
13                 dp[i][j] = dp[i - 1][j];
14                 if (S.charAt(i - 1) == T.charAt(j - 1)) 
15                     dp[i][j] += dp[i - 1][j - 1];
16             }
17         }
18      
19         return dp[S.length()][T.length()];
20     }

 Reference:http://blog.csdn.net/abcbc/article/details/8978146