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LeetCode OJ - Distinct Subsequence

2024-07-02 22:08:51   226人阅读

这道题采用动态规划，可是我一开始没有想到。后来参考了discuss中前辈的代码和思路，才想通的。 方法二是因为每一步只和上一步的内容相关，所以可以只用O（n）的空间复杂度。

下面是AC代码：

 1 /**
 2      * Solution DP
 3      * we keep a m*n matrix and scanning through string T,
 4      * p[i][j] means the number of distinct subsequence of S(0...j) equal to T(0...i) 
 5      * p[i][j] =  p[i][j-1]   discard s[j]
 6      *         +  0           if s[j] != t[i]
 7      *         +  p[i-1][j-1] if s[j] == t[i]
 8      * @param S
 9      * @param T
10      * @return
11      */
12     public int numDistinct1(String S, String T){
13         if(S.length()<T.length())
14             return 0;
15         int[][] p = new int[T.length()][S.length()];
16         //for convenient
17         char[] Sc = S.toCharArray();
18         char[] Tc = T.toCharArray();
19         
20         p[0][0] = Tc[0] == Sc[0] ? 1:0;
21         
22         for(int i=1;i<S.length();i++)
23             p[0][i] = p[0][i-1]+(Tc[0] == Sc[i]?1:0);
24         
25         for(int j = 1;j<T.length();j++)
26             p[j][0] = 0;
27         
28         for(int i=1;i<T.length();i++){
29             for(int j=1;j<S.length();j++){
30                 p[i][j] = p[i][j-1]+(Tc[i] == Sc[j] ? p[i-1][j-1]:0);
31             }
32         }
33         return p[T.length()-1][S.length()-1];
34     }
35     /**
36      * O(n) space complexity solution
37      * @param S
38      * @param T
39      * @return
40      */
41     public int numDistinct2(String S,String T){
42         if(S.length()<T.length())
43             return 0;
44         int[] p = new int[S.length()];
45         //for convenient
46         char[] Sc = S.toCharArray();
47         char[] Tc = T.toCharArray();
48        
49         p[0] = Tc[0] == Sc[0] ? 1:0;
50         
51         for(int i=1;i<S.length();i++)
52             p[i] = p[i-1]+(Tc[0] == Sc[i]?1:0);
53         
54         for(int i=1;i<T.length();i++){
55             for(int j=S.length()-1;j>=0;j--){
56                 p[j] = p[j] + (Tc[i] == Sc[j]? p[j-1] :0);
57             }
58         }
59         return p[S.length()-1];
60        
61     }

 


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