feat: add solutions to lc problems: No.2907,2921

solution/0100-0199/0114.Flatten Binary Tree to Linked List/README.md

@@ -59,7 +59,7 @@
 
 因此，对于当前节点，如果其左子节点不为空，我们找到左子树的最右节点，作为前驱节点，然后将当前节点的右子节点赋给前驱节点的右子节点。然后将当前节点的左子节点赋给当前节点的右子节点，并将当前节点的左子节点置为空。然后将当前节点的右子节点作为下一个节点，继续处理，直至所有节点处理完毕。
 
-时间复杂度 $O(n)$，空间复杂度 O(1)$。其中 $n$ 是树中节点的个数。
+时间复杂度 $O(n)$，其中 $n$ 是树中节点的个数。空间复杂度 O(1)$。
 
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solution/0100-0199/0115.Distinct Subsequences/README_EN.md

@@ -44,6 +44,32 @@ As shown below, there are 5 ways you can generate "bag" from s.
 
 ## Solutions
 
+**Solution 1: Dynamic Programming**
+
+We define $f[i][j]$ as the number of schemes where the first $i$ characters of string $s$ form the first $j$ characters of string $t$. Initially, $f[i][0]=1$ for all $i \in [0,m]$.
+
+When $i > 0$, we consider the calculation of $f[i][j]$:
+
+-   When $s[i-1] \ne t[j-1]$, we cannot select $s[i-1]$, so $f[i][j]=f[i-1][j]$;
+-   Otherwise, we can select $s[i-1]$, so $f[i][j]=f[i-1][j-1]$.
+
+Therefore, we have the following state transition equation:
+
+$$
+f[i][j]=\left\{
+\begin{aligned}
+&f[i-1][j], &s[i-1] \ne t[j-1] \\
+&f[i-1][j-1]+f[i-1][j], &s[i-1]=t[j-1]
+\end{aligned}
+\right.
+$$
+
+The final answer is $f[m][n]$, where $m$ and $n$ are the lengths of strings $s$ and $t$ respectively.
+
+The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$.
+
+We notice that the calculation of $f[i][j]$ is only related to $f[i-1][..]$. Therefore, we can optimize the first dimension, reducing the space complexity to $O(n)$.
+
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 ### **Python3**