9.10 叠箱子，不是很理解，建议做几道dp的leetcode题目

本文介绍了一种使用动态规划方法来寻找最高堆叠箱组合的算法实现。该算法通过递归方式检查每个箱子是否可以放置在另一个箱子之上，并记录最大高度的堆叠方案。

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You have a stack of n boxes, with widths w., heights l\ and depths dr The boxes
cannot be rotated and can only be stacked on top of one another if each box in the
stack is strictly larger than the box above it in width, height, and depth. Implement
a method to build the tallest stack possible, where the heigh t of a stack is the sum of
the heights of each box. 答案cc150 165页

public ArrayList<Box> createStackDP(Box[] boxes, Box bottom,

2 HashMap<Box, ArrayList<Box» stack_map) {
3 if (bottom != null && stack_map.containsKey(bottom)) {
4 return stack_map.get(bottom);
5 }
6
int max_height = 0;
8 ArrayList<Box> max_stack = null;
9 for (int i = 0; i < boxes.length; i++) {
10 if (boxes[i].canBeAbove(bottom)) {
11 ArrayList<Box> new_stack =
12 createStackDP(boxes, boxes[i], stackjnap);
13 int new_height = stackHeight(new_stack);
14 if (newjieight > max_height) {
15 max_stack = new_stack;
16 max_height = newjieight;
17 }
18 }
19 }
20
21 if (max_stack == null) max_stack = new ArrayList<Box>();
22 if (bottom != null) max_stack.add(0, bottom);
23 stackjnap.put(bottomj max_stack);
24
25 return (ArrayList<Box>)max_stack.clone();
26 }