5353
5454<!-- 这里可写通用的实现逻辑 -->
5555
56- 两次 dp，` dp1[i][j] ` 表示 i ~ j 的子串是否是回文，可以参考 [ 5. 最长回文子串] ( ../../solution/0000-0099/0005.Longest%20Palindromic%20Substring/README.md ) 。` dp2[i] ` 表示以 i 结尾的子串最少需要分割几次，如果本来就是回文串（` dp[0][i] == true ` ）就不需要分割，否则枚举分割点 ` j `
56+ ** 方法一：动态规划**
57+
58+ 我们先预处理得到字符串 $s$ 的每一个子串 $s[ i..j] $ 是否为回文串，记录在二维数组 $g[ i] [ j ] $ 中，其中 $g[ i] [ j ] $ 表示子串 $s[ i..j] $ 是否为回文串。
59+
60+ 接下来，我们定义 $f[ i] $ 表示字符串 $s[ 0..i-1] $ 的最少分割次数，初始时 $f[ i] =i$。
61+
62+ 接下来，我们考虑 $f[ i] $ 如何进行状态转移。我们可以枚举上一个分割点 $j$，如果子串 $s[ j..i] $ 是一个回文串，那么 $f[ i] $ 就可以从 $f[ j] $ 转移而来。如果 $j=0$，那么说明 $s[ 0..i] $ 本身就是一个回文串，此时不需要进行分割，即 $f[ i] =0$。因此，状态转移方程如下：
63+
64+ $$
65+ f[i]=\min_{0\leq j \leq i}\begin{cases} f[j-1]+1, & \text{if}\ g[j][i]=\text{True} \\ 0, & \text{if}\ g[0][i]=\text{True} \end{cases}
66+ $$
67+
68+ 答案即为 $f[ n] $，其中 $n$ 是字符串 $s$ 的长度。
69+
70+ 时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 是字符串 $s$ 的长度。
5771
5872<!-- tabs:start -->
5973
6579class Solution :
6680 def minCut (self s : str ) -> int :
6781 n = len (s)
68- dp1 = [[False ] * n for _ in range (n)]
82+ g = [[True ] * n for _ in range (n)]
6983 for i in range (n - 1 , - 1 , - 1 ):
70- for j in range (i, n):
71- dp1[i][j] = s[i] == s[j] and (j - i < 3 or dp1[i + 1 ][j - 1 ])
72- dp2 = [0 ] * n
73- for i in range (n):
74- if not dp1[0 ][i]:
75- dp2[i] = i
76- for j in range (1 , i + 1 ):
77- if dp1[j][i]:
78- dp2[i] = min (dp2[i], dp2[j - 1 ] + 1 )
79- return dp2[- 1 ]
84+ for j in range (i + 1 , n):
85+ g[i][j] = s[i] == s[j] and g[i + 1 ][j - 1 ]
86+ f = list (range (n))
87+ for i in range (1 , n):
88+ for j in range (i + 1 ):
89+ if g[j][i]:
90+ f[i] = min (f[i], 1 + f[j - 1 ] if j else 0 )
91+ return f[- 1 ]
8092```
8193
8294### ** Java**
@@ -87,54 +99,90 @@ class Solution:
8799class Solution {
88100 public int minCut (String s ) {
89101 int n = s. length();
90- boolean [][] dp1 = new boolean [n][n];
91- for (int i = n - 1 ; i >= 0 ; i-- ) {
92- for (int j = i; j < n; j++ ) {
93- dp1[i][j] = s. charAt(i) == s. charAt(j) && (j - i < 3 || dp1[i + 1 ][j - 1 ]);
102+ boolean [][] g = new boolean [n][n];
103+ for (var row : g) {
104+ Arrays . fill(row, true );
105+ }
106+ for (int i = n - 1 ; i >= 0 ; -- i) {
107+ for (int j = i + 1 ; j < n; ++ j) {
108+ g[i][j] = s. charAt(i) == s. charAt(j) && g[i + 1 ][j - 1 ];
94109 }
95110 }
96- int [] dp2 = new int [n];
97- for (int i = 0 ; i < n; i ++ ) {
98- if ( ! dp1[ 0 ][i]) {
99- dp2[i] = i;
100- for (int j = 1 ; j <= i; j ++ ) {
101- if (dp1[j][i] ) {
102- dp2[i] = Math . min(dp2 [i], dp2[j - 1 ] + 1 );
103- }
111+ int [] f = new int [n];
112+ for (int i = 0 ; i < n; ++ i ) {
113+ f[i] = i;
114+ }
115+ for (int i = 1 ; i < n; ++ i ) {
116+ for ( int j = 0 ; j <= i; ++ j ) {
117+ if (g[j] [i]) {
118+ f[i] = Math . min(f[i], j > 0 ? 1 + f[j - 1 ] : 0 );
104119 }
105120 }
106121 }
107- return dp2 [n - 1 ];
122+ return f [n - 1 ];
108123 }
109124}
110125```
111126
127+ ### ** C++**
128+
129+ ``` cpp
130+ class Solution {
131+ public:
132+ int minCut(string s) {
133+ int n = s.size();
134+ bool g[ n] [ n ] ;
135+ memset(g, true, sizeof(g));
136+ for (int i = n - 1; ~ i; --i) {
137+ for (int j = i + 1; j < n; ++j) {
138+ g[ i] [ j ] = s[ i] == s[ j] && g[ i + 1] [ j - 1 ] ;
139+ }
140+ }
141+ int f[ n] ;
142+ iota(f, f + n, 0);
143+ for (int i = 1; i < n; ++i) {
144+ for (int j = 0; j <= i; ++j) {
145+ if (g[ j] [ i ] ) {
146+ f[ i] = min(f[ i] , j ? 1 + f[ j - 1] : 0);
147+ }
148+ }
149+ }
150+ return f[ n - 1] ;
151+ }
152+ };
153+ ```
154+
112155### **Go**
113156
114157```go
115158func minCut(s string) int {
116159 n := len(s)
117- dp1 := make ([][]bool , n)
118- for i := 0 ; i < n; i++ {
119- dp1[i] = make ([]bool , n)
160+ g := make([][]bool, n)
161+ f := make([]int, n)
162+ for i := range g {
163+ g[i] = make([]bool, n)
164+ f[i] = i
165+ for j := range g[i] {
166+ g[i][j] = true
167+ }
120168 }
121169 for i := n - 1; i >= 0; i-- {
122- for j := i; j < n; j++ {
123- dp1 [i][j] = s[i] == s[j] && (j-i < 3 || dp1 [i+1 ][j-1 ])
170+ for j := i + 1 ; j < n; j++ {
171+ g [i][j] = s[i] == s[j] && g [i+1][j-1]
124172 }
125173 }
126- dp2 := make ([] int , n)
127- for i := 0 ; i < n; i ++ {
128- if !dp1[ 0 ][i] {
129- dp2[i] = i
130- for j := 1 ; j <= i; j++ {
131- if dp1[j][i] {
132- dp2 [i] = min (dp2 [i], dp2 [j-1 ]+1 )
174+ for i := 1; i < n; i++ {
175+ for j := 0; j <= i; j ++ {
176+ if g[j ][i] {
177+ if j == 0 {
178+ f[i] = 0
179+ } else {
180+ f [i] = min(f [i], f [j-1]+1)
133181 }
134182 }
135183 }
136184 }
137- return dp2 [n-1 ]
185+ return f [n-1]
138186}
139187
140188func min(x, y int) int {
@@ -145,33 +193,62 @@ func min(x, y int) int {
145193}
146194```
147195
148- ### ** C++**
196+ ### ** TypeScript**
197+
198+ ``` ts
199+ function minCut(s : string ): number {
200+ const = s .length ;
201+ const : boolean [][] = Array (n )
202+ .fill (0 )
203+ .map (() => Array (n ).fill (true ));
204+ for (let i = n - 1 ; ~ i ; -- i ) {
205+ for (let j = i + 1 ; j < n ; ++ j ) {
206+ g [i ][j ] = s [i ] === s [j ] && g [i + 1 ][j - 1 ];
207+ }
208+ }
209+ const : number [] = Array (n )
210+ .fill (0 )
211+ .map ((_ , i ) => i );
212+ for (let i = 1 ; i < n ; ++ i ) {
213+ for (let j = 0 ; j <= i ; ++ j ) {
214+ if (g [j ][i ]) {
215+ f [i ] = Math .min (f [i ], j ? 1 + f [j - 1 ] : 0 );
216+ }
217+ }
218+ }
219+ return f [n - 1 ];
220+ }
221+ ```
149222
150- ``` cpp
151- class Solution {
152- public:
153- int minCut(string s) {
154- int n = s.size();
155- vector<vector<bool >> dp1(n, vector<bool >(n));
223+ ### ** C#**
224+
225+ ``` cs
226+ public class Solution {
227+ public int MinCut (string s ) {
228+ int n = s .Length ;
229+ bool [,] g = new bool [n ,n ];
230+ int [] f = new int [n ];
231+ for (int i = 0 ; i < n ; ++ i ) {
232+ f [i ] = i ;
233+ for (int j = 0 ; j < n ; ++ j ) {
234+ g [i ,j ] = true ;
235+ }
236+ }
156237 for (int i = n - 1 ; i >= 0 ; -- i ) {
157- for (int j = i; j < n; ++j) {
158- dp1 [ i ] [ j ] = s[ i] == s[ j] && (j - i < 3 || dp1 [ i + 1] [ j - 1 ] ) ;
238+ for (int j = i + 1 ; j < n ; ++ j ) {
239+ g [ i , j ] = s [i ] == s [j ] && g [i + 1 , j - 1 ];
159240 }
160241 }
161- vector<int > dp2(n);
162- for (int i = 0; i < n; ++i) {
163- if (!dp1[ 0] [ i ] ) {
164- dp2[ i] = i;
165- for (int j = 1; j <= i; ++j) {
166- if (dp1[ j] [ i ] ) {
167- dp2[ i] = min(dp2[ i] , dp2[ j - 1] + 1);
168- }
242+ for (int i = 1 ; i < n ; ++ i ) {
243+ for (int j = 0 ; j <= i ; ++ j ) {
244+ if (g [j ,i ]) {
245+ f [i ] = Math .Min (f [i ], j > 0 ? 1 + f [j - 1 ] : 0 );
169246 }
170247 }
171248 }
172- return dp2 [ n - 1] ;
249+ return f [n - 1 ];
173250 }
174- };
251+ }
175252```
176253
177254### ** ...**
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