5454
5555<!-- 这里可写通用的实现逻辑 -->
5656
57- 动态规划，` dp[i][j] ` 表示 ` s[:i] ` 的子序列中 ` t[:j] ` 的出现次数
57+ ** 方法一：动态规划**
58+
59+ 我们定义 $f[ i] [ j ] $ 表示字符串 $s$ 的前 $i$ 个字符中，子序列构成字符串 $t$ 的前 $j$ 个字符的方案数。初始时 $f[ i] [ 0 ] =1$，其中 $i \in [ 0,m] $。
60+
61+ 当 $i \gt 0$ 时，考虑 $f[ i] [ j ] $ 的计算：
62+
63+ - 当 $s[ i-1] \ne t[ j-1] $ 时，不能选取 $s[ i-1] $，因此 $f[ i] [ j ] =f[ i-1] [ j ] $；
64+ - 否则，可以选取 $s[ i-1] $，此时 $f[ i] [ j ] =f[ i-1] [ j-1 ] $。
65+
66+ 因此我们有如下的状态转移方程：
67+
68+ $$
69+ f[i][j]=\left\{
70+ \begin{aligned}
71+ &f[i-1][j], &s[i-1] \ne t[j-1] \\
72+ &f[i-1][j-1]+f[i-1][j], &s[i-1]=t[j-1]
73+ \end{aligned}
74+ \right.
75+ $$
76+
77+ 最终的答案即为 $f[ m] [ n ] $，其中 $m$ 和 $n$ 分别是字符串 $s$ 和 $t$ 的长度。
78+
79+ 时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。
80+
81+ 我们注意到 $f[ i] [ j ] $ 的计算只和 $f[ i-1] [ .. ] $ 有关，因此，我们可以优化掉第一维，这样空间复杂度可以降低到 $O(n)$。
5882
5983<!-- tabs:start -->
6084
6690class Solution :
6791 def numDistinct (self s : str , t : str ) -> int :
6892 m, n = len (s), len (t)
69- dp = [[0 ] * (n + 1 ) for _ in range (m + 1 )]
93+ f = [[0 ] * (n + 1 ) for _ in range (m + 1 )]
7094 for i in range (m + 1 ):
71- dp[i][0 ] = 1
72- for i in range (1 , m + 1 ):
73- for j in range (1 , n + 1 ):
74- dp[i][j] = dp[i - 1 ][j]
75- if s[i - 1 ] == t[j - 1 ]:
76- dp[i][j] += dp[i - 1 ][j - 1 ]
77- return dp[m][n]
95+ f[i][0 ] = 1
96+ for i, a in enumerate (s, 1 ):
97+ for j, b in enumerate (t, 1 ):
98+ f[i][j] = f[i - 1 ][j]
99+ if a == b:
100+ f[i][j] += f[i - 1 ][j - 1 ]
101+ return f[m][n]
102+ ```
103+
104+ ``` python
105+ class Solution :
106+ def numDistinct (self s : str , t : str ) -> int :
107+ n = len (t)
108+ f = [1 ] + [0 ] * n
109+ for a in s:
110+ for j in range (n, 0 , - 1 ):
111+ if a == t[j - 1 ]:
112+ f[j] += f[j - 1 ]
113+ return f[n]
78114```
79115
80116### ** Java**
@@ -84,69 +120,168 @@ class Solution:
84120``` java
85121class Solution {
86122 public int numDistinct (String s , String t ) {
87- int m = s. length();
88- int n = t. length();
89- int [][] dp = new int [m + 1 ][n + 1 ];
90- for (int i = 0 ; i <= m; i++ ) {
91- dp[i][0 ] = 1 ;
123+ int m = s. length(), n = t. length();
124+ int [][] f = new int [m + 1 ][n + 1 ];
125+ for (int i = 0 ; i < m + 1 ; ++ i) {
126+ f[i][0 ] = 1 ;
92127 }
93- for (int i = 1 ; i <= m; i ++ ) {
94- for (int j = 1 ; j <= n; j ++ ) {
95- dp [i][j] = dp [i - 1 ][j];
128+ for (int i = 1 ; i < m + 1 ; ++ i ) {
129+ for (int j = 1 ; j < n + 1 ; ++ j ) {
130+ f [i][j] = f [i - 1 ][j];
96131 if (s. charAt(i - 1 ) == t. charAt(j - 1 )) {
97- dp[i][j] += dp[i - 1 ][j - 1 ];
132+ f[i][j] += f[i - 1 ][j - 1 ];
133+ }
134+ }
135+ }
136+ return f[m][n];
137+ }
138+ }
139+ ```
140+
141+ ``` java
142+ class Solution {
143+ public int numDistinct (String s , String t ) {
144+ int n = t. length();
145+ int [] f = new int [n + 1 ];
146+ f[0 ] = 1 ;
147+ for (char a : s. toCharArray()) {
148+ for (int j = n; j > 0 ; -- j) {
149+ char b = t. charAt(j - 1 );
150+ if (a == b) {
151+ f[j] += f[j - 1 ];
98152 }
99153 }
100154 }
101- return dp[m] [n];
155+ return f [n];
102156 }
103157}
104158```
105159
160+ ### ** C++**
161+
162+ ``` cpp
163+ class Solution {
164+ public:
165+ int numDistinct(string s, string t) {
166+ int m = s.size(), n = t.size();
167+ unsigned long long f[ m + 1] [ n + 1 ] ;
168+ memset(f, 0, sizeof(f));
169+ for (int i = 0; i < m + 1; ++i) {
170+ f[ i] [ 0 ] = 1;
171+ }
172+ for (int i = 1; i < m + 1; ++i) {
173+ for (int j = 1; j < n + 1; ++j) {
174+ f[ i] [ j ] = f[ i - 1] [ j ] ;
175+ if (s[ i - 1] == t[ j - 1] ) {
176+ f[ i] [ j ] += f[ i - 1] [ j - 1 ] ;
177+ }
178+ }
179+ }
180+ return f[ m] [ n ] ;
181+ }
182+ };
183+ ```
184+
185+ ```cpp
186+ class Solution {
187+ public:
188+ int numDistinct(string s, string t) {
189+ int n = t.size();
190+ unsigned long long f[n + 1];
191+ memset(f, 0, sizeof(f));
192+ f[0] = 1;
193+ for (char& a : s) {
194+ for (int j = n; j; --j) {
195+ char b = t[j - 1];
196+ if (a == b) {
197+ f[j] += f[j - 1];
198+ }
199+ }
200+ }
201+ return f[n];
202+ }
203+ };
204+ ```
205+
106206### ** Go**
107207
108208``` go
109209func numDistinct (s string , t string ) int {
110210 m , n := len (s), len (t)
111- dp := make ([][]int , m+1 )
211+ f := make ([][]int , m+1 )
212+ for i := range f {
213+ f[i] = make ([]int , n+1 )
214+ }
112215 for i := 0 ; i <= m; i++ {
113- dp[i] = make ([]int , n+1 )
114- dp[i][0 ] = 1
216+ f[i][0 ] = 1
115217 }
116218 for i := 1 ; i <= m; i++ {
117219 for j := 1 ; j <= n; j++ {
118- dp [i][j] = dp [i-1 ][j]
220+ f [i][j] = f [i-1 ][j]
119221 if s[i-1 ] == t[j-1 ] {
120- dp [i][j] += dp [i-1 ][j-1 ]
222+ f [i][j] += f [i-1 ][j-1 ]
121223 }
122224 }
123225 }
124- return dp [m][n]
226+ return f [m][n]
125227}
126228```
127229
128- ### ** C++**
230+ ``` go
231+ func numDistinct (s string , t string ) int {
232+ n := len (t)
233+ f := make ([]int , n+1 )
234+ f[0 ] = 1
235+ for _ , a := range s {
236+ for j := n; j > 0 ; j-- {
237+ if b := t[j-1 ]; byte (a) == b {
238+ f[j] += f[j-1 ]
239+ }
240+ }
241+ }
242+ return f[n]
243+ }
244+ ```
129245
130- ``` cpp
131- class Solution {
132- public:
133- int numDistinct(string s, string t) {
134- int m = s.size(), n = t.size();
135- vector<vector<unsigned long long >> dp(m + 1, vector<unsigned long long >(n + 1));
136- for (int i = 0; i <= m; ++i) {
137- dp[ i] [ 0 ] = 1;
246+ ### ** TypeScript**
247+
248+ ``` ts
249+ function numDistinct(s : string , t : string ): number {
250+ const = s .length ;
251+ const = t .length ;
252+ const : number [][] = new Array (m + 1 )
253+ .fill (0 )
254+ .map (() => new Array (n + 1 ).fill (0 ));
255+ for (let i = 0 ; i <= m ; ++ i ) {
256+ f [i ][0 ] = 1 ;
257+ }
258+ for (let i = 1 ; i <= m ; ++ i ) {
259+ for (let j = 1 ; j <= n ; ++ j ) {
260+ f [i ][j ] = f [i - 1 ][j ];
261+ if (s [i - 1 ] === t [j - 1 ]) {
262+ f [i ][j ] += f [i - 1 ][j - 1 ];
263+ }
138264 }
139- for (int i = 1; i <= m; ++i) {
140- for (int j = 1; j <= n; ++j) {
141- dp[ i] [ j ] = dp[ i - 1] [ j ] ;
142- if (s[ i - 1] == t[ j - 1] ) {
143- dp[ i] [ j ] += dp[ i - 1] [ j - 1 ] ;
144- }
265+ }
266+ return f [m ][n ];
267+ }
268+ ```
269+
270+ ``` ts
271+ function numDistinct(s : string , t : string ): number {
272+ const = t .length ;
273+ const : number [] = new Array (n + 1 ).fill (0 );
274+ f [0 ] = 1 ;
275+ for (const of s ) {
276+ for (let j = n ; j ; -- j ) {
277+ const = t [j - 1 ];
278+ if (a === b ) {
279+ f [j ] += f [j - 1 ];
145280 }
146281 }
147- return dp[ m] [ n ] ;
148282 }
149- };
283+ return f [n ];
284+ }
150285```
151286
152287### ** ...**
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