5757
5858<!-- 这里可写通用的实现逻辑 -->
5959
60+ ** 方法一：动态规划**
61+
62+ 我们用 $dp[ i] $ 表示摆放前 $i$ 本书所需要的书架的最小高度，初始时 $dp[ 0] =0$。
63+
64+ 遍历每一本书 $books[ i-1] $，把这本书放在书架新的一层，那么有 $dp[ i] =dp[ i-1] +h$。我们还可以将这本书往前的书放在与这本书放在同一层，尽可能减少书架的高度，那么有 $dp[ i] =min(dp[ i] , dp[ j-1] +h)$。其中 $h$ 是最后一层的书的最大高度。
65+
66+ 时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 表示数组 ` books ` 的长度。
67+
6068<!-- tabs:start -->
6169
6270### ** Python3**
@@ -68,15 +76,13 @@ class Solution:
6876 def minHeightShelves (self books : List[List[int ]], shelfWidth : int ) -> int :
6977 n = len (books)
7078 dp = [0 ] * (n + 1 )
71- dp[1 ] = books[0 ][1 ]
72- for i in range (2 , n + 1 ):
73- dp[i] = books[i - 1 ][1 ] + dp[i - 1 ]
74- w, h = books[i - 1 ][0 ], books[i - 1 ][1 ]
79+ for i, (w, h) in enumerate (books, 1 ):
80+ dp[i] = dp[i - 1 ] + h
7581 for j in range (i - 1 , 0 , - 1 ):
7682 w += books[j - 1 ][0 ]
7783 if w > shelfWidth:
7884 break
79- h = max (books[j - 1 ][1 ], h )
85+ h = max (h, books[j - 1 ][1 ])
8086 dp[i] = min (dp[i], dp[j - 1 ] + h)
8187 return dp[n]
8288```
@@ -90,11 +96,10 @@ class Solution {
9096 public int minHeightShelves (int [][] books , int shelfWidth ) {
9197 int n = books. length;
9298 int [] dp = new int [n + 1 ];
93- dp[1 ] = books[0 ][1 ];
94- for (int i = 2 ; i <= n; i++ ) {
95- dp[i] = dp[i - 1 ] + books[i - 1 ][1 ];
99+ for (int i = 1 ; i <= n; ++ i) {
96100 int w = books[i - 1 ][0 ], h = books[i - 1 ][1 ];
97- for (int j = i - 1 ; j > 0 ; j-- ) {
101+ dp[i] = dp[i - 1 ] + h;
102+ for (int j = i - 1 ; j > 0 ; -- j) {
98103 w += books[j - 1 ][0 ];
99104 if (w > shelfWidth) {
100105 break ;
@@ -108,6 +113,65 @@ class Solution {
108113}
109114```
110115
116+ ### ** C++**
117+
118+ ``` cpp
119+ class Solution {
120+ public:
121+ int minHeightShelves(vector<vector<int >>& books, int shelfWidth) {
122+ int n = books.size();
123+ vector<int > dp(n + 1);
124+ for (int i = 1; i <= n; ++i) {
125+ int w = books[ i - 1] [ 0 ] , h = books[ i - 1] [ 1 ] ;
126+ dp[ i] = dp[ i - 1] + h;
127+ for (int j = i - 1; j > 0; --j) {
128+ w += books[ j - 1] [ 0 ] ;
129+ if (w > shelfWidth) break;
130+ h = max(h, books[ j - 1] [ 1 ] );
131+ dp[ i] = min(dp[ i] , dp[ j - 1] + h);
132+ }
133+ }
134+ return dp[ n] ;
135+ }
136+ };
137+ ```
138+
139+ ### **Go**
140+
141+ ```go
142+ func minHeightShelves(books [][]int, shelfWidth int) int {
143+ n := len(books)
144+ dp := make([]int, n+1)
145+ for i := 1; i <= n; i++ {
146+ w, h := books[i-1][0], books[i-1][1]
147+ dp[i] = dp[i-1] + h
148+ for j := i - 1; j > 0; j-- {
149+ w += books[j-1][0]
150+ if w > shelfWidth {
151+ break
152+ }
153+ h = max(h, books[j-1][1])
154+ dp[i] = min(dp[i], dp[j-1]+h)
155+ }
156+ }
157+ return dp[n]
158+ }
159+
160+ func max(a, b int) int {
161+ if a > b {
162+ return a
163+ }
164+ return b
165+ }
166+
167+ func min(a, b int) int {
168+ if a < b {
169+ return a
170+ }
171+ return b
172+ }
173+ ```
174+
111175### ** ...**
112176
113177```
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