8888
8989<!-- solution:start -->
9090
91- ### 方法一:记忆化搜索
91+ ### 方法一:前缀和 + 记忆化搜索
92+ 93+ 我们用一个数组 $\textit{nums}$ 记录每个单词的长度,数组的长度记为 $n$。然后我们定义一个长度为 $n + 1$ 的前缀和数组 $\textit{s},ドル其中 $\textit{s}[ i] $ 表示前 $i$ 个单词的长度之和。
94+ 95+ 接下来,我们设计一个函数 $\textit{dfs}(i),ドル表示从第 $i$ 个单词开始分隔句子的最小成本。那么答案为 $\textit{dfs}(0)$。
96+ 97+ 函数 $\textit{dfs}(i)$ 的执行过程如下:
98+ 99+ - 如果从第 $i$ 个单词开始到最后一个单词的长度之和加上单词之间的空格数小于等于 $k,ドル那么这些单词可以放在最后一行,成本为 0ドル$。
100+ - 否则,我们枚举下一个开始分隔的单词的位置 $j,ドル使得从第 $i$ 个单词到第 $j-1$ 个单词的长度之和加上单词之间的空格数小于等于 $k$。那么 $\textit{dfs}(j)$ 表示从第 $j$ 个单词开始分隔句子的最小成本,而 $(k - m)^2$ 表示将第 $i$ 个单词到第 $j-1$ 个单词放在一行的成本,其中 $m$ 表示从第 $i$ 个单词到第 $j-1$ 个单词的长度之和加上单词之间的空格数。我们枚举所有的 $j,ドル取最小值即可。
101+ 102+ 答案即为 $\textit{dfs}(0)$。
103+ 104+ 为了避免重复计算,我们可以使用记忆化搜索。
105+ 106+ 时间复杂度 $O(n^2),ドル空间复杂度 $O(n)$。其中 $n$ 为单词的个数。
92107
93108<!-- tabs:start -->
94109
@@ -98,59 +113,56 @@ tags:
98113class Solution :
99114 def minimumCost (self sentence : str , k : int ) -> int :
100115 @cache
101- def dfs (i ) :
102- if s[- 1 ] - s[i] + n - i - 1 <= k:
116+ def dfs (i : int ) -> int :
117+ if s[n ] - s[i] + n - i - 1 <= k:
103118 return 0
104- ans, j = inf, i + 1
105- while j < n and (t := s[j] - s[i] + j - i - 1 ) <= k:
106- ans = min (ans, (k - t) ** 2 + dfs(j))
119+ ans = inf
120+ j = i + 1
121+ while j < n and (m := s[j] - s[i] + j - i - 1 ) <= k:
122+ ans = min (ans, dfs(j) + (k - m) ** 2 )
107123 j += 1
108124 return ans
109125
110- t = [len (w ) for w in sentence.split()]
111- n = len (t )
112- s = list (accumulate(t , initial = 0 ))
126+ nums = [len (s ) for s in sentence.split()]
127+ n = len (nums )
128+ s = list (accumulate(nums , initial = 0 ))
113129 return dfs(0 )
114130```
115131
116132#### Java
117133
118134``` java
119135class Solution {
120- private static final int INF = Integer . MAX_VALUE
121- private int [] memo;
136+ private Integer [] f;
122137 private int [] s;
138+ private int k;
123139 private int n;
124140
125141 public int minimumCost (String sentence , int k ) {
142+ this . k = k;
126143 String [] words = sentence. split(" "
127144 n = words. length;
145+ f = new Integer [n];
128146 s = new int [n + 1 ];
129147 for (int i = 0 ; i < n; ++ i) {
130148 s[i + 1 ] = s[i] + words[i]. length();
131149 }
132- memo = new int [n];
133- Arrays . fill(memo, INF );
134- return dfs(0 , k);
150+ return dfs(0 );
135151 }
136152
137- private int dfs (int i , int k ) {
138- if (memo[i] != INF ) {
139- return memo[i];
140- }
153+ private int dfs (int i ) {
141154 if (s[n] - s[i] + n - i - 1 <= k) {
142- memo[i] = 0 ;
143155 return 0 ;
144156 }
145- int ans = INF ;
146- for (int j = i + 1 ; j < n; ++ j) {
147- int t = s[j] - s[i] + j - i - 1 ;
148- if (t <= k) {
149- ans = Math . min(ans, (k - t) * (k - t) + dfs(j, k));
150- }
157+ if (f[i] != null ) {
158+ return f[i];
159+ }
160+ int ans = Integer . MAX_VALUE
161+ for (int j = i + 1 ; j < n && s[j] - s[i] + j - i - 1 <= k; ++ j) {
162+ int m = s[j] - s[i] + j - i - 1 ;
163+ ans = Math . min(ans, dfs(j) + (k - m) * (k - m));
151164 }
152- memo[i] = ans;
153- return ans;
165+ return f[i] = ans;
154166 }
155167}
156168```
@@ -160,34 +172,31 @@ class Solution {
160172``` cpp
161173class Solution {
162174public:
163- const int inf = INT_MAX;
164- int n;
165- 166175 int minimumCost(string sentence, int k) {
167- istringstream is(sentence);
168- vector<string> words;
169- string word;
170- while (is >> word) words.push_back(word);
171- n = words.size();
172- vector<int> s(n + 1);
173- for (int i = 0; i < n; ++i) s[i + 1] = s[i] + words[i].size();
174- vector<int> memo(n, inf);
175- return dfs(0, k, s, memo);
176- }
177- 178- int dfs (int i, int k, vector<int >& s, vector<int >& memo) {
179- if (memo[ i] != inf) return memo[ i] ;
180- if (s[ n] - s[ i] + n - i - 1 <= k) {
181- memo[ i] = 0;
182- return 0;
183- }
184- int ans = inf;
185- for (int j = i + 1; j < n; ++j) {
186- int t = s[ j] - s[ i] + j - i - 1;
187- if (t <= k) ans = min(ans, (k - t) * (k - t) + dfs(j, k, s, memo));
176+ istringstream iss(sentence);
177+ vector<int > s = {0};
178+ string w;
179+ while (iss >> w) {
180+ s.push_back(s.back() + w.size());
188181 }
189- memo[ i] = ans;
190- return ans;
182+ int n = s.size() - 1;
183+ int f[ n] ;
184+ memset(f, -1, sizeof(f));
185+ auto dfs = [ &] (auto&& dfs, int i) -> int {
186+ if (s[ n] - s[ i] + n - i - 1 <= k) {
187+ return 0;
188+ }
189+ if (f[ i] != -1) {
190+ return f[ i] ;
191+ }
192+ int ans = INT_MAX;
193+ for (int j = i + 1; j < n && s[ j] - s[ i] + j - i - 1 <= k; ++j) {
194+ int m = s[ j] - s[ i] + j - i - 1;
195+ ans = min(ans, dfs(dfs, j) + (k - m) * (k - m));
196+ }
197+ return f[ i] = ans;
198+ };
199+ return dfs(dfs, 0);
191200 }
192201};
193202```
@@ -196,40 +205,63 @@ public:
196205
197206```go
198207func minimumCost(sentence string, k int) int {
199- words := strings.Split(sentence, " ")
200- n := len(words)
201- inf := math.MaxInt32
202- s := make([]int, n+1)
203- for i, word := range words {
204- s[i+1] = s[i] + len(word)
208+ s := []int{0}
209+ for _, w := range strings.Split(sentence, " ") {
210+ s = append(s, s[len(s)-1]+len(w))
205211 }
206- memo := make([]int, n)
207- for i := range memo {
208- memo[i] = inf
212+ n := len(s) - 1
213+ f := make([]int, n)
214+ for i := range f {
215+ f[i] = -1
209216 }
210217 var dfs func(int) int
211218 dfs = func(i int) int {
212- if memo[i] != inf {
213- return memo[i]
214- }
215219 if s[n]-s[i]+n-i-1 <= k {
216- memo[i] = 0
217220 return 0
218221 }
219- ans := inf
220- for j := i + 1; j < n; j++ {
221- t := s[j] - s[i] + j - i - 1
222- if t <= k {
223- ans = min(ans, (k-t)*(k-t)+dfs(j))
224- }
222+ if f[i] != -1 {
223+ return f[i]
225224 }
226- memo[i] = ans
225+ ans := math.MaxInt32
226+ for j := i + 1; j < n && s[j]-s[i]+j-i-1 <= k; j++ {
227+ m := s[j] - s[i] + j - i - 1
228+ ans = min(ans, dfs(j)+(k-m)*(k-m))
229+ }
230+ f[i] = ans
227231 return ans
228232 }
229233 return dfs(0)
230234}
231235```
232236
237+ #### TypeScript
238+ 239+ ``` ts
240+ function minimumCost(sentence : string , k : number ): number {
241+ const : number [] = [0 ];
242+ for (const of sentence .split (' '
243+ s .push (s .at (- 1 )! + w .length );
244+ }
245+ const = s .length - 1 ;
246+ const : number [] = Array (n ).fill (- 1 );
247+ const = (i : number ): number => {
248+ if (s [n ] - s [i ] + n - i - 1 <= k ) {
249+ return 0 ;
250+ }
251+ if (f [i ] !== - 1 ) {
252+ return f [i ];
253+ }
254+ let ans = Infinity ;
255+ for (let j = i + 1 ; j < n && s [j ] - s [i ] + j - i - 1 <= k ; ++ j ) {
256+ const = s [j ] - s [i ] + j - i - 1 ;
257+ ans = Math .min (ans , dfs (j ) + (k - m ) ** 2 );
258+ }
259+ return (f [i ] = ans );
260+ };
261+ return dfs (0 );
262+ }
263+ ```
264+ 233265<!-- tabs: end -->
234266
235267<!-- solution: end -->
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