1 class Solution
 2 {
 3     public:
 4         int distinctSubseqII(string S)
 5         {
 6             int n = S.size();
 7             int mod = 1000000007;
 8             int h[2002],last[2002],dp[2002];
 9             char a[2002];
10             memset(h,0,sizeof(h));
11             memset(last,0,sizeof(last));
 
12             memset(dp,0,sizeof(dp));
13             for(int i=1; i<=n; ++i)
14             {
15                 a[i]=S[i-1];
16                 last[i]=h[a[i]];
17                 h[a[i]]=i;
18             }
19             for(int i=1; i<=n; ++i)
20             {
21                 if(!last[i])
 
22                     dp[i]=dp[i-1]*2+1;
23                 else
24                     dp[i]=dp[i-1]+dp[i-1]-dp[last[i]-1];
25                 if(dp[i]<0)
26                     dp[i]+=mod;
27                 dp[i]%=mod;
28             }
29             return dp[n];
30         }
31 };

代碼借鑒：https://blog.csdn.net/icefox_zhx/article/details/77254400

Leetcode-940 Distinct Subsequences II(不同的子序列 II)