Well, a dynamic programming problem. Let's first define its state dp[i][j] to be the number of distinct subsequences of t[0..i - 1] in s[0..j - 1]. Then we have the following state equations:

1. General case 1: dp[i][j] = dp[i][j - 1] if t[i - 1] != s[j - 1];
2. General case 2: dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1] if t[i - 1] == s[j - 1];
3. Boundary case 1: dp[0][j] = 1 for all j;
4. Boundary case 2: dp[i][0] = 0 for all positive i.

Now let's give brief explanations to the four equations above.

1. If t[i - 1] != s[j - 1], the distinct subsequences will not include s[j - 1] and thus all the number of distinct subsequences will simply be those in s[0..j - 2], which corresponds to dp[i][j - 1];
2. If t[i - 1] == s[j - 1], the number of distinct subsequences include two parts: those withs[j - 1] and those without;
3. An empty string will have exactly one subsequence in any string :-)
4. Non-empty string will have no subsequences in an empty string.

Putting these together, we will have the following simple codes (just like translation :-)):

1 class Solution { 2 public: 3     int numDistinct(string s, string t) { 4         int m = t.length(), n = s.length(); 5         vector
     
      
       > dp(m + 1, vector
       
         (n + 1, 0)); 6         for (int j = 0; j <= n; j++) dp[0][j] = 1; 7         for (int j = 1; j <= n; j++) 8             for (int i = 1; i <= m; i++) 9                 dp[i][j] = dp[i][j - 1] + (t[i - 1] == s[j - 1] ? dp[i - 1][j - 1] : 0);10         return dp[m][n];11     }12 };
       
      
     

Notice that we keep the whole m*n matrix simply for dp[i - 1][j - 1]. So we can simply store that value in a single variable and further optimize the space complexity. The final code is as follows.

1 class Solution { 2 public: 3     int numDistinct(string s, string t) { 4         int m = t.length(), n = s.length(); 5         vector
     
       cur(m + 1, 0); 6         cur[0] = 1; 7         for (int j = 1; j <= n; j++) {  8             int pre = 1; 9             for (int i = 1; i <= m; i++) {10                 int temp = cur[i];11                 cur[i] = cur[i] + (t[i - 1] == s[j - 1] ? pre : 0);12                 pre = temp;13             }14         }15         return cur[m];16     }17 };