@@ -56,7 +56,23 @@ coins = 3*1*5 + 3*5*8 + 1*3*8 + 1*8*1 = 167</pre>
5656
5757<!-- solution:start -->
5858
59- ### 方法一
59+ ### 方法一:动态规划
60+ 61+ 我们记数组 $nums$ 的长度为 $n$。根据题目描述,我们可以在数组 $nums$ 的左右两端各添加一个 1ドル,ドル记为 $arr$。
62+ 63+ 然后,我们定义 $f[ i] [ j ] $ 表示戳破区间 $[ i, j] $ 内的所有气球能得到的最多硬币数,那么答案即为 $f[ 0] [ n+1 ] $。
64+ 65+ 对于 $f[ i] [ j ] ,ドル我们枚举区间 $[ i, j] $ 内的所有位置 $k,ドル假设 $k$ 是最后一个戳破的气球,那么我们可以得到如下状态转移方程:
66+ 67+ $$
68+ f[i][j] = \max(f[i][j], f[i][k] + f[k][j] + arr[i] \times arr[k] \times arr[j])
69+ $$
70+ 71+ 在实现上,由于 $f[ i] [ j ] $ 的状态转移方程中涉及到 $f[ i] [ k ] $ 和 $f[ k] [ j ] ,ドル其中 $i < k < j,ドル因此我们需要从大到小地遍历 $i,ドル从小到大地遍历 $j,ドル这样才能保证当计算 $f[ i] [ j ] $ 时 $f[ i] [ k ] $ 和 $f[ k] [ j ] $ 已经被计算出来。
72+ 73+ 最后,我们返回 $f[ 0] [ n+1 ] $ 即可。
74+ 75+ 时间复杂度 $O(n^3),ドル空间复杂度 $O(n^2)$。其中 $n$ 为数组 $nums$ 的长度。
6076
6177<!-- tabs:start -->
6278
@@ -65,40 +81,35 @@ coins = 3*1*5 + 3*5*8 + 1*3*8 + 1*8*1 = 167</pre>
6581``` python
6682class Solution :
6783 def maxCoins (self nums : List[int ]) -> int :
68- nums = [1 ] + nums + [1 ]
6984 n = len (nums)
70- dp = [[ 0 ] * n for _ in range (n) ]
71- for l in range (2 , n):
72- for i in range (n - l ):
73- j = i + l
85+ arr = [1 ] + nums + [ 1 ]
86+ f = [[ 0 ] * (n + 2 ) for _ in range (n + 2 )]
87+ for i in range (n - 1 , - 1 , - 1 ):
88+ for j in range (i + 2 , n + 2 ):
7489 for k in range (i + 1 , j):
75- dp[i][j] = max (
76- dp[i][j], dp[i][k] + dp[k][j] + nums[i] * nums[k] * nums[j]
77- )
78- return dp[0 ][- 1 ]
90+ f[i][j] = max (f[i][j], f[i][k] + f[k][j] + arr[i] * arr[k] * arr[j])
91+ return f[0 ][- 1 ]
7992```
8093
8194#### Java
8295
8396``` java
8497class Solution {
8598 public int maxCoins (int [] nums ) {
86- int [] vals = new int [nums. length + 2 ];
87- vals[0 ] = 1 ;
88- vals[vals. length - 1 ] = 1 ;
89- System . arraycopy(nums, 0 , vals, 1 , nums. length);
90- int n = vals. length;
91- int [][] dp = new int [n][n];
92- for (int l = 2 ; l < n; ++ l) {
93- for (int i = 0 ; i + l < n; ++ i) {
94- int j = i + l;
95- for (int k = i + 1 ; k < j; ++ k) {
96- dp[i][j]
97- = Math . max(dp[i][j], dp[i][k] + dp[k][j] + vals[i] * vals[k] * vals[j]);
99+ int n = nums. length;
100+ int [] arr = new int [n + 2 ];
101+ arr[0 ] = 1 ;
102+ arr[n + 1 ] = 1 ;
103+ System . arraycopy(nums, 0 , arr, 1 , n);
104+ int [][] f = new int [n + 2 ][n + 2 ];
105+ for (int i = n - 1 ; i >= 0 ; i-- ) {
106+ for (int j = i + 2 ; j <= n + 1 ; j++ ) {
107+ for (int k = i + 1 ; k < j; k++ ) {
108+ f[i][j] = Math . max(f[i][j], f[i][k] + f[k][j] + arr[i] * arr[k] * arr[j]);
98109 }
99110 }
100111 }
101- return dp [0 ][n - 1 ];
112+ return f [0 ][n + 1 ];
102113 }
103114}
104115```
@@ -109,19 +120,21 @@ class Solution {
109120class Solution {
110121public:
111122 int maxCoins(vector<int >& nums) {
112- nums.insert(nums.begin(), 1);
113- nums.push_back(1);
114123 int n = nums.size();
115- vector<vector<int >> dp(n, vector<int >(n));
116- for (int l = 2; l < n; ++l) {
117- for (int i = 0; i + l < n; ++i) {
118- int j = i + l;
124+ vector<int > arr(n + 2, 1);
125+ for (int i = 0; i < n; ++i) {
126+ arr[ i + 1] = nums[ i] ;
127+ }
128+ 129+ vector<vector<int>> f(n + 2, vector<int>(n + 2, 0));
130+ for (int i = n - 1; i >= 0; --i) {
131+ for (int j = i + 2; j <= n + 1; ++j) {
119132 for (int k = i + 1; k < j; ++k) {
120- dp [ i] [ j ] = max(dp [ i] [ j ] , dp [ i] [ k ] + dp [ k] [ j ] + nums [ i] * nums [ k] * nums [ j] );
133+ f [i][j] = max(f [i][j], f [i][k] + f [k][j] + arr [i] * arr [k] * arr [j]);
121134 }
122135 }
123136 }
124- return dp [ 0] [ n - 1 ] ;
137+ return f [0 ][n + 1 ];
125138 }
126139};
127140```
@@ -130,44 +143,72 @@ public:
130143
131144``` go
132145func maxCoins (nums []int ) int {
133- vals := make([]int, len(nums)+2)
134- for i := 0; i < len(nums); i++ {
135- vals[i+1] = nums[i]
136- }
137- n := len(vals)
138- vals[0], vals[n-1] = 1, 1
139- dp := make([][]int, n)
140- for i := 0; i < n; i++ {
141- dp[i] = make([]int, n)
142- }
143- for l := 2; l < n; l++ {
144- for i := 0; i+l < n; i++ {
145- j := i + l
146- for k := i + 1; k < j; k++ {
147- dp[i][j] = max(dp[i][j], dp[i][k]+dp[k][j]+vals[i]*vals[k]*vals[j])
148- }
149- }
150- }
151- return dp[0][n-1]
146+ n := len (nums)
147+ arr := make ([]int , n+2 )
148+ arr[0 ] = 1
149+ arr[n+1 ] = 1
150+ copy (arr[1 :], nums)
151+ 152+ f := make ([][]int , n+2 )
153+ for i := range f {
154+ f[i] = make ([]int , n+2 )
155+ }
156+ 157+ for i := n - 1 ; i >= 0 ; i-- {
158+ for j := i + 2 ; j <= n+1 ; j++ {
159+ for k := i + 1 ; k < j; k++ {
160+ f[i][j] = max (f[i][j], f[i][k] + f[k][j] + arr[i]*arr[k]*arr[j])
161+ }
162+ }
163+ }
164+ 165+ return f[0 ][n+1 ]
152166}
153167```
154168
155169#### TypeScript
156170
157171``` ts
158172function maxCoins(nums : number []): number {
159- let n = nums .length ;
160- let dp = Array .from ({ length: n + 1 }, v => new Array (n + 2 ).fill (0 ));
161- nums .unshift (1 );
162- nums .push (1 );
163- for (let i = n - 1 ; i >= 0 ; -- i ) {
164- for (let j = i + 2 ; j < n + 2 ; ++ j ) {
165- for (let k = i + 1 ; k < j ; ++ k ) {
166- dp [i ][j ] = Math .max (nums [i ] * nums [k ] * nums [j ] + dp [i ][k ] + dp [k ][j ], dp [i ][j ]);
173+ const = nums .length ;
174+ const = Array (n + 2 ).fill (1 );
175+ for (let i = 0 ; i < n ; i ++ ) {
176+ arr [i + 1 ] = nums [i ];
177+ }
178+ 179+ const : number [][] = Array .from ({ length: n + 2 }, () => Array (n + 2 ).fill (0 ));
180+ for (let i = n - 1 ; i >= 0 ; i -- ) {
181+ for (let j = i + 2 ; j <= n + 1 ; j ++ ) {
182+ for (let k = i + 1 ; k < j ; k ++ ) {
183+ f [i ][j ] = Math .max (f [i ][j ], f [i ][k ] + f [k ][j ] + arr [i ] * arr [k ] * arr [j ]);
184+ }
185+ }
186+ }
187+ return f [0 ][n + 1 ];
188+ }
189+ ```
190+ 191+ #### Rust
192+ 193+ ``` rust
194+ impl Solution {
195+ pub fn max_coins (nums : Vec <i32 >) -> i32 {
196+ let n = nums . len ();
197+ let mut arr = vec! [1 ; n + 2 ];
198+ for i in 0 .. n {
199+ arr [i + 1 ] = nums [i ];
200+ }
201+ 202+ let mut f = vec! [vec! [0 ; n + 2 ]; n + 2 ];
203+ for i in (0 .. n ). rev () {
204+ for j in i + 2 .. n + 2 {
205+ for k in i + 1 .. j {
206+ f [i ][j ] = f [i ][j ]. max (f [i ][k ] + f [k ][j ] + arr [i ] * arr [k ] * arr [j ]);
207+ }
167208 }
168209 }
210+ f [0 ][n + 1 ]
169211 }
170- return dp [0 ][n + 1 ];
171212}
172213```
173214
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