4848
4949<!-- 这里可写通用的实现逻辑 -->
5050
51- 题目可以转换为 ` 0-1 ` 背包问题,在 k 个字符串中选出一些字符串(每个字符串只能使用一次),并且满足字符串最多包含 m 个 0 和 n 个 1,求满足此条件的字符串的最大长度(字符串个数)。
51+ ** 方法一:动态规划**
52+ 53+ 我们定义 $f[ i] [ j ] [ k] $ 表示在前 $i$ 个字符串中,使用 $j$ 个 0 和 $k$ 个 1 的情况下最多可以得到的字符串数量。初始时 $f[ i] [ j ] [ k] =0,ドル答案为 $f[ sz] [ m ] [ n] ,ドル其中 $sz$ 是数组 $strs$ 的长度。
54+ 55+ 对于 $f[ i] [ j ] [ k] ,ドル我们有两种决策:
56+ 57+ - 不选第 $i$ 个字符串,此时 $f[ i] [ j ] [ k] =f[ i-1] [ j ] [ k] $;
58+ - 选第 $i$ 个字符串,此时 $f[ i] [ j ] [ k] =f[ i-1] [ j-a ] [ k-b] +1,ドル其中 $a$ 和 $b$ 分别是第 $i$ 个字符串中 0ドル$ 和 1ドル$ 的数量。
59+ 60+ 我们取两种决策中的最大值,即可得到 $f[ i] [ j ] [ k] $ 的值。
61+ 62+ 最终的答案即为 $f[ sz] [ m ] [ n] $。
63+ 64+ 时间复杂度 $O(sz \times m \times n),ドル空间复杂度 $O(sz \times m \times n)$。其中 $sz$ 是数组 $strs$ 的长度;而 $m$ 和 $n$ 分别是字符串中 0ドル$ 和 1ドル$ 的数量上限。
65+ 66+ 我们注意到 $f[ i] [ j ] [ k] $ 的计算只和 $f[ i-1] [ j ] [ k] $ 以及 $f[ i-1] [ j-a ] [ k-b] $ 有关,因此我们可以去掉第一维,将空间复杂度优化到 $O(m \times n)$。
5267
5368<!-- tabs:start -->
5469
5974``` python
6075class Solution :
6176 def findMaxForm (self strs : List[str ], m : int , n : int ) -> int :
62- l = len (strs)
63- dp = [[[0 ] * (n + 1 ) for i in range (m + 1 )] for j in range (l)]
64- t = [(s.count(' 0' ' 1' for s in strs]
65- n0, n1 = t[0 ]
66- for j in range (m + 1 ):
67- for k in range (n + 1 ):
68- if n0 <= j and n1 <= k:
69- dp[0 ][j][k] = 1
70- 71- for i in range (1 , l):
72- n0, n1 = t[i]
77+ sz = len (strs)
78+ f = [[[0 ] * (n + 1 ) for _ in range (m + 1 )] for _ in range (sz + 1 )]
79+ for i, s in enumerate (strs, 1 ):
80+ a, b = s.count(" 0" " 1"
7381 for j in range (m + 1 ):
7482 for k in range (n + 1 ):
75- dp[i][j][k] = dp[i - 1 ][j][k]
76- if n0 <= j and n1 <= k:
77- dp[i][j][k] = max (dp[i][j][k], dp[i - 1 ][j - n0][k - n1] + 1 )
78- 79- return dp[- 1 ][- 1 ][- 1 ]
83+ f[i][j][k] = f[i - 1 ][j][k]
84+ if j >= a and k >= b:
85+ f[i][j][k] = max (f[i][j][k], f[i - 1 ][j - a][k - b] + 1 )
86+ return f[sz][m][n]
8087```
8188
82- 空间优化:
83- 8489``` python
8590class Solution :
8691 def findMaxForm (self strs : List[str ], m : int , n : int ) -> int :
87- dp = [[0 ] * (n + 1 ) for _ in range (m + 1 )]
88- t = [(s.count(' 0' ' 1' for s in strs]
89- for k in range (len (strs)):
90- n0, n1 = t[k]
91- for i in range (m, n0 - 1 , - 1 ):
92- for j in range (n, n1 - 1 , - 1 ):
93- dp[i][j] = max (dp[i][j], dp[i - n0][j - n1] + 1 )
94- return dp[- 1 ][- 1 ]
92+ f = [[0 ] * (n + 1 ) for _ in range (m + 1 )]
93+ for s in strs:
94+ a, b = s.count(" 0" " 1"
95+ for i in range (m, a - 1 , - 1 ):
96+ for j in range (n, b - 1 , - 1 ):
97+ f[i][j] = max (f[i][j], f[i - a][j - b] + 1 )
98+ return f[m][n]
9599```
96100
97101### ** Java**
@@ -100,24 +104,56 @@ class Solution:
100104
101105``` java
102106class Solution {
103- public : int findMaxForm (vector <string >& strs , int m , int n ) {
104- vector< vector< int >> dp(m + 1 , vector< int > (n + 1 ));
105- for (auto s : strs) {
106- vector< int > t = count(s);
107- for (int i = m; i >= t[0 ]; -- i)
108- for (int j = n; j >= t[1 ]; -- j)
109- dp[i][j] = max(dp[i][j], dp[i - t[0 ]][j - t[1 ]] + 1 );
107+ public int findMaxForm (String [] strs , int m , int n ) {
108+ int sz = strs. length;
109+ int [][][] f = new int [sz + 1 ][m + 1 ][n + 1 ];
110+ for (int i = 1 ; i <= sz; ++ i) {
111+ int [] cnt = count(strs[i - 1 ]);
112+ for (int j = 0 ; j <= m; ++ j) {
113+ for (int k = 0 ; k <= n; ++ k) {
114+ f[i][j][k] = f[i - 1 ][j][k];
115+ if (j >= cnt[0 ] && k >= cnt[1 ]) {
116+ f[i][j][k] = Math . max(f[i][j][k], f[i - 1 ][j - cnt[0 ]][k - cnt[1 ]] + 1 );
117+ }
118+ }
119+ }
110120 }
111- return dp [m][n];
121+ return f[sz] [m][n];
112122 }
113123
114- vector<int > count (string s ) {
115- int n0 = 0 ;
116- for (char c : s)
117- if (c == ' 0' ++ n0;
118- return {n0, (int ) s. size() - n0};
124+ private int [] count (String s ) {
125+ int [] cnt = new int [2 ];
126+ for (int i = 0 ; i < s. length(); ++ i) {
127+ ++ cnt[s. charAt(i) - ' 0'
128+ }
129+ return cnt;
119130 }
120- };
131+ }
132+ ```
133+ 134+ ``` java
135+ class Solution {
136+ public int findMaxForm (String [] strs , int m , int n ) {
137+ int [][] f = new int [m + 1 ][n + 1 ];
138+ for (String s : strs) {
139+ int [] cnt = count(s);
140+ for (int i = m; i >= cnt[0 ]; -- i) {
141+ for (int j = n; j >= cnt[1 ]; -- j) {
142+ f[i][j] = Math . max(f[i][j], f[i - cnt[0 ]][j - cnt[1 ]] + 1 );
143+ }
144+ }
145+ }
146+ return f[m][n];
147+ }
148+ 149+ private int [] count (String s ) {
150+ int [] cnt = new int [2 ];
151+ for (int i = 0 ; i < s. length(); ++ i) {
152+ ++ cnt[s. charAt(i) - ' 0'
153+ }
154+ return cnt;
155+ }
156+ }
121157```
122158
123159### ** C++**
@@ -126,21 +162,50 @@ class Solution {
126162class Solution {
127163public:
128164 int findMaxForm(vector<string >& strs, int m, int n) {
129- vector<vector<int >> dp(m + 1, vector<int >(n + 1));
130- for (int k = 0; k < strs.size(); ++k) {
131- vector<int > t = count(strs[ k] );
132- for (int i = m; i >= t[ 0] ; --i)
133- for (int j = n; j >= t[ 1] ; --j)
134- dp[ i] [ j ] = max(dp[ i] [ j ] , dp[ i - t[ 0]] [ j - t[ 1]] + 1);
165+ int sz = strs.size();
166+ int f[ sz + 1] [ m + 1 ] [ n + 1] ;
167+ memset(f, 0, sizeof(f));
168+ for (int i = 1; i <= sz; ++i) {
169+ auto [ a, b] = count(strs[ i - 1] );
170+ for (int j = 0; j <= m; ++j) {
171+ for (int k = 0; k <= n; ++k) {
172+ f[ i] [ j ] [ k] = f[ i - 1] [ j ] [ k] ;
173+ if (j >= a && k >= b) {
174+ f[ i] [ j ] [ k] = max(f[ i] [ j ] [ k] , f[ i - 1] [ j - a ] [ k - b] + 1);
175+ }
176+ }
177+ }
178+ }
179+ return f[ sz] [ m ] [ n] ;
180+ }
181+ 182+ pair<int, int> count(string& s) {
183+ int a = count_if(s.begin(), s.end(), [](char c) { return c == '0'; });
184+ return {a, s.size() - a};
185+ }
186+ };
187+ ```
188+ 189+ ``` cpp
190+ class Solution {
191+ public:
192+ int findMaxForm(vector<string >& strs, int m, int n) {
193+ int f[ m + 1] [ n + 1 ] ;
194+ memset(f, 0, sizeof(f));
195+ for (auto& s : strs) {
196+ auto [ a, b] = count(s);
197+ for (int i = m; i >= a; --i) {
198+ for (int j = n; j >= b; --j) {
199+ f[ i] [ j ] = max(f[ i] [ j ] , f[ i - a] [ j - b ] + 1);
200+ }
201+ }
135202 }
136- return dp [ m] [ n ] ;
203+ return f [ m] [ n ] ;
137204 }
138205
139- vector<int> count(string s) {
140- int n0 = 0;
141- for (char c : s)
142- if (c == '0') ++n0;
143- return {n0, (int) s.size() - n0};
206+ pair<int, int> count(string& s) {
207+ int a = count_if(s.begin(), s.end(), [](char c) { return c == '0'; });
208+ return {a, s.size() - a};
144209 }
145210};
146211```
@@ -149,29 +214,61 @@ public:
149214
150215``` go
151216func findMaxForm (strs []string , m int , n int ) int {
152- dp := make ([][]int , m+1 )
153- for i := 0 ; i < m+1 ; i++ {
154- dp[i] = make ([]int , n+1 )
217+ sz := len (strs)
218+ f := make ([][][]int , sz+1 )
219+ for i := range f {
220+ f[i] = make ([][]int , m+1 )
221+ for j := range f[i] {
222+ f[i][j] = make ([]int , n+1 )
223+ }
155224 }
156- for _ , s := range strs {
157- t := count (s)
158- for i := m; i >= t[0 ]; i-- {
159- for j := n; j >= t[1 ]; j-- {
160- dp[i][j] = max (dp[i][j], dp[i-t[0 ]][j-t[1 ]]+1 )
225+ for i := 1 ; i <= sz; i++ {
226+ a , b := count (strs[i-1 ])
227+ for j := 0 ; j <= m; j++ {
228+ for k := 0 ; k <= n; k++ {
229+ f[i][j][k] = f[i-1 ][j][k]
230+ if j >= a && k >= b {
231+ f[i][j][k] = max (f[i][j][k], f[i-1 ][j-a][k-b]+1 )
232+ }
161233 }
162234 }
163235 }
164- return dp[m][n]
236+ return f[sz][m][n]
237+ }
238+ 239+ func count (s string ) (int , int ) {
240+ a := strings.Count (s, " 0"
241+ return a, len (s) - a
242+ }
243+ 244+ func max (a , b int ) int {
245+ if a > b {
246+ return a
247+ }
248+ return b
165249}
250+ ```
166251
167- func count (s string ) []int {
168- n0 := 0
169- for i := 0 ; i < len (s); i++ {
170- if s[i] == ' 0'
171- n0++
252+ ``` go
253+ func findMaxForm (strs []string , m int , n int ) int {
254+ f := make ([][]int , m+1 )
255+ for i := range f {
256+ f[i] = make ([]int , n+1 )
257+ }
258+ for _ , s := range strs {
259+ a , b := count (s)
260+ for j := m; j >= a; j-- {
261+ for k := n; k >= b; k-- {
262+ f[j][k] = max (f[j][k], f[j-a][k-b]+1 )
263+ }
172264 }
173265 }
174- return []int {n0, len (s) - n0}
266+ return f[m][n]
267+ }
268+ 269+ func count (s string ) (int , int ) {
270+ a := strings.Count (s, " 0"
271+ return a, len (s) - a
175272}
176273
177274func max (a , b int ) int {
@@ -182,6 +279,58 @@ func max(a, b int) int {
182279}
183280```
184281
282+ ### ** TypeScript**
283+ 284+ ``` ts
285+ function findMaxForm(strs : string [], m : number , n : number ): number {
286+ const = strs .length ;
287+ const = Array .from ({ length: sz + 1 }, () =>
288+ Array .from ({ length: m + 1 }, () => Array .from ({ length: n + 1 }, () => 0 )),
289+ );
290+ const = (s : string ): [number , number ] => {
291+ let a = 0 ;
292+ for (const of s ) {
293+ a += c === ' 0' ? 1 : 0 ;
294+ }
295+ return [a , s .length - a ];
296+ };
297+ for (let i = 1 ; i <= sz ; ++ i ) {
298+ const = count (strs [i - 1 ]);
299+ for (let j = 0 ; j <= m ; ++ j ) {
300+ for (let k = 0 ; k <= n ; ++ k ) {
301+ f [i ][j ][k ] = f [i - 1 ][j ][k ];
302+ if (j >= a && k >= b ) {
303+ f [i ][j ][k ] = Math .max (f [i ][j ][k ], f [i - 1 ][j - a ][k - b ] + 1 );
304+ }
305+ }
306+ }
307+ }
308+ return f [sz ][m ][n ];
309+ }
310+ ```
311+ 312+ ``` ts
313+ function findMaxForm(strs : string [], m : number , n : number ): number {
314+ const = Array .from ({ length: m + 1 }, () => Array .from ({ length: n + 1 }, () => 0 ));
315+ const = (s : string ): [number , number ] => {
316+ let a = 0 ;
317+ for (const of s ) {
318+ a += c === ' 0' ? 1 : 0 ;
319+ }
320+ return [a , s .length - a ];
321+ };
322+ for (const of strs ) {
323+ const = count (s );
324+ for (let i = m ; i >= a ; -- i ) {
325+ for (let j = n ; j >= b ; -- j ) {
326+ f [i ][j ] = Math .max (f [i ][j ], f [i - a ][j - b ] + 1 );
327+ }
328+ }
329+ }
330+ return f [m ][n ];
331+ }
332+ ```
333+ 185334### ** ...**
186335
187336```
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