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里看到的一句話：

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

 所以這種類型題可以多往DP思考思考。

 首先設置動態規划數組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，那么dp[0][j]都是0；

 如果T串為空，那么dp[i][j]都是1，因為空串為是任何字符串的字串。

 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，如果字符串S[i-1]和T[j-1]（dp是從1開始計數，字符串是從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]。

 

代碼如下：

 
           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