Problem:

Given a string S, count the number of distinct, non-empty subsequences of S .

Since the result may be large, return the answer modulo 10^9 + 7.

Example 1:

Input: "abc"
Output: 7
Explanation: The 7 distinct subsequences are "a", "b", "c", "ab", "ac", "bc", and "abc".

Example 2:

Input: "aba"
Output: 6
Explanation: The 6 distinct subsequences are "a", "b", "ab", "ba", "aa" and "aba".

Example 3:

Input: "aaa"
Output: 3
Explanation: The 3 distinct subsequences are "a", "aa" and "aaa".

Note:

1. S contains only lowercase letters.
2. 1 <= S.length <= 2000

Solution:

　　這道題說白了一點都不難，問題是我沒想到。拿到題目的第一反映是動態規劃，但看到S的長度只有2000，我思路一直限制在了提高維度這一方向上，結果答案只是O（n）的時間複雜度。這裡用了一個last陣列記錄了所有小寫字母結尾的子序列數目，從頭遍歷字串S，在所有子序列後面加上S[i]可以得到新的子序列，然後加上S[i]本身就是以S[i]結尾的新的子序列個數。程式碼非常簡單。

Code:

 1 class Solution {
 2 public:
 3     int distinctSubseqII(string S) {
 4         int
 
 m = S.size();
 5         int result = 0;
 6         vector<int> last(26,0);
 7         for(int i = 0;i != m;++i){
 8             int sum = 0;
 9             for(int i = 0;i != 26;++i)
10                 sum = (sum+last[i])%1000000007;
11             last[S[i]-'a'] = sum+1;
12         }
13         for(int i = 0;i != 26;++i)
14             result = (result+last[i])%1000000007;
15         return result;
16     }
17 };