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feat: add solutions to lc problem: No.0115
No.0115.Distinct Subsequences
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‎solution/0100-0199/0115.Distinct Subsequences/README.md

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@@ -52,7 +52,25 @@
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<!-- 这里可写通用的实现逻辑 -->
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动态规划，`dp[i][j]` 表示 `s[:i]` 的子序列中 `t[:j]` 的出现次数
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**方法一：动态规划**
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定义 `dp[i][j]` 表示 `s[0..i]` 的子序列中 `t[0..j]` 出现的个数。初始时 `dp[i][0]=1`，表示空串是任意字符串的子序列。答案为 `dp[m][n]`。
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当 `s[i] == t[j]` 时，`dp[i][j] = dp[i-1][j-1] + dp[i-1][j]`，即 `s[0..i]` 的子序列中 `t[0..j]` 出现的个数等于 `s[0..i-1]` 的子序列中 `t[0..j-1]` 出现的个数加上 `s[0..i-1]` 的子序列中 `t[0..j]` 出现的个数。
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当 `s[i] != t[j]` 时，`dp[i][j] = dp[i-1][j]`，即 `s[0..i]` 的子序列中 `t[0..j]` 出现的个数等于 `s[0..i-1]` 的子序列中 `t[0..j]` 出现的个数。
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因此，可以得到状态转移方程：
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$$
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dp[i][j]=
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\begin{cases}
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dp[i-1][j-1]+dp[i-1][j], & s[i]=t[j] \\
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dp[i-1][j], & s[i]\neq t[j]
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\end{cases}
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$$
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时间复杂度 $O(m\times n)$，空间复杂度 $O(m\times n)$。其中 $m$, $n$ 分别是字符串 `s` 和 `t` 的长度。
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<!-- tabs:start -->
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@@ -82,15 +100,14 @@ class Solution:
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```java
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class Solution {
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public int numDistinct(String s, String t) {
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int m = s.length();
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int n = t.length();
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int m = s.length(), n = t.length();
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int[][] dp = new int[m + 1][n + 1];
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for (int i = 0; i <= m; i++) {
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for (int i = 0; i <= m; ++i) {
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dp[i][0] = 1;
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}
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for (int i = 1; i <= m; i++) {
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for (int j = 1; j <= n; j++) {
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dp[i][j] = dp[i - 1][j];
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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dp[i][j] += dp[i - 1][j];
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if (s.charAt(i - 1) == t.charAt(j - 1)) {
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dp[i][j] += dp[i - 1][j - 1];
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}
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};
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```
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### **TypeScript**
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```ts
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function numDistinct(s: string, t: string): number {
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const m = s.length;
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const n = t.length;
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const dp: number[][] = new Array(m + 1)
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.fill(0)
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.map(() => new Array(n + 1).fill(0));
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for (let i = 0; i <= m; ++i) {
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dp[i][0] = 1;
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}
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for (let i = 1; i <= m; ++i) {
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for (let j = 1; j <= n; ++j) {
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dp[i][j] = dp[i - 1][j];
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if (s.charAt(i - 1) === t.charAt(j - 1)) {
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dp[i][j] += dp[i - 1][j - 1];
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}
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}
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}
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return dp[m][n];
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}
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```
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### **...**
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```

‎solution/0100-0199/0115.Distinct Subsequences/README_EN.md

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@@ -70,15 +70,14 @@ class Solution:
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```java
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class Solution {
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public int numDistinct(String s, String t) {
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int m = s.length();
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int n = t.length();
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int m = s.length(), n = t.length();
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int[][] dp = new int[m + 1][n + 1];
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for (int i = 0; i <= m; i++) {
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for (int i = 0; i <= m; ++i) {
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dp[i][0] = 1;
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}
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for (int i = 1; i <= m; i++) {
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for (int j = 1; j <= n; j++) {
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dp[i][j] = dp[i - 1][j];
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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dp[i][j] += dp[i - 1][j];
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if (s.charAt(i - 1) == t.charAt(j - 1)) {
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dp[i][j] += dp[i - 1][j - 1];
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}
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};
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```
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### **TypeScript**
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```ts
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function numDistinct(s: string, t: string): number {
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const m = s.length;
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const n = t.length;
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const dp: number[][] = new Array(m + 1)
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.fill(0)
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.map(() => new Array(n + 1).fill(0));
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for (let i = 0; i <= m; ++i) {
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dp[i][0] = 1;
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}
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for (let i = 1; i <= m; ++i) {
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for (let j = 1; j <= n; ++j) {
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dp[i][j] = dp[i - 1][j];
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if (s.charAt(i - 1) === t.charAt(j - 1)) {
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dp[i][j] += dp[i - 1][j - 1];
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}
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}
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}
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return dp[m][n];
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}
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```
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### **...**
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```

‎solution/0100-0199/0115.Distinct Subsequences/Solution.cpp

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}
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return dp[m][n];
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}
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};
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};

‎solution/0100-0199/0115.Distinct Subsequences/Solution.java

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@@ -1,19 +1,18 @@
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class Solution {
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public int numDistinct(String s, String t) {
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int m = s.length();
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int n = t.length();
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int m = s.length(), n = t.length();
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int[][] dp = new int[m + 1][n + 1];
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for (int i = 0; i <= m; i++) {
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for (int i = 0; i <= m; ++i) {
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dp[i][0] = 1;
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}
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for (int i = 1; i <= m; i++) {
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for (int j = 1; j <= n; j++) {
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dp[i][j] = dp[i - 1][j];
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for (int i = 1; i <= m; ++i) {
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for (int j = 1; j <= n; ++j) {
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dp[i][j] += dp[i - 1][j];
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if (s.charAt(i - 1) == t.charAt(j - 1)) {
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dp[i][j] += dp[i - 1][j - 1];
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}
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}
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}
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return dp[m][n];
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}
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}
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}

‎solution/0100-0199/0115.Distinct Subsequences/Solution.py

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class Solution:
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def numDistinct(self, s: str, t: str) -> int:
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m, n = len(s), len(t)
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dp = [[0] * n + 1 for _ in range(m + 1)]
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dp = [[0] * (n + 1) for _ in range(m + 1)]
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for i in range(m + 1):
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dp[i][0] = 1
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for i in range(1, m + 1):
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for j in range(1, n + 1):
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dp[i][j] = dp[i - 1][j]
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if s[i - 1] == s[j - 1]:
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if s[i - 1] == t[j - 1]:
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dp[i][j] += dp[i - 1][j - 1]
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return dp[m][n]

‎solution/0100-0199/0115.Distinct Subsequences/Solution.ts

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function numDistinct(s: string, t: string): number {
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const m = s.length;
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const n = t.length;
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const dp: number[][] = new Array(m + 1)
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.fill(0)
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.map(() => new Array(n + 1).fill(0));
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for (let i = 0; i <= m; ++i) {
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dp[i][0] = 1;
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}
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for (let i = 1; i <= m; ++i) {
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for (let j = 1; j <= n; ++j) {
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dp[i][j] = dp[i - 1][j];
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if (s.charAt(i - 1) === t.charAt(j - 1)) {
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dp[i][j] += dp[i - 1][j - 1];
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}
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}
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}
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return dp[m][n];
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}

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