Description

We have a sequence of books: the i-th book has thickness books[i][0] and height books[i][1].

We want to place these books in order onto bookcase shelves that have total width shelf_width.

We choose some of the books to place on this shelf (such that the sum of their thickness is <= shelf_width), then build another level of shelf of the bookcase so that the total height of the bookcase has increased by the maximum height of the books we just put down. We repeat this process until there are no more books to place.

Note again that at each step of the above process, the order of the books we place is the same order as the given sequence of books. For example, if we have an ordered list of 5 books, we might place the first and second book onto the first shelf, the third book on the second shelf, and the fourth and fifth book on the last shelf.

Return the minimum possible height that the total bookshelf can be after placing shelves in this manner.

[leetcode] 1105. Filling Bookcase Shelves_参考文献

Example 1:

Input: books = [[1,1],[2,3],[2,3],[1,1],[1,1],[1,1],[1,2]], shelf_width = 4
Output: 6
Explanation:
The sum of the heights of the 3 shelves are 1 + 3 + 2 = 6.
Notice that book number 2 does not have to be on the first shelf.

Constraints:

  • 1 <= books.length <= 1000
  • 1 <= books[i][0] <= shelf_width <= 1000
  • 1 <= books[i][1] <= 1000

分析

题目的意思是：给你一个数组，数组里面子数组第一个数值表示书的宽度，第二个数值表示书的高度。用 dp[i] 表示放置前 i 本书所需要的书架最小高度，初始值 dp[0] = 0，其他为最大值 inf。遍历每一本书，把当前这本书作为书架最后一层的最后一本书，将这本书之前的书向后调整，看看是否可以减少之前的书架高度。

dp[i] = min(dp[i] , dp[j - 1] + h)

其中 j 表示最后一层所能容下书籍的索引，h 表示最后一层最大高度。

代码

class Solution:
    def minHeightShelves(self, books: List[List[int]], shelf_width: int) -> int:
        n=len(books)
        dp=[float('inf')]*(n+1)
        // 遍历求摆放前i本书的最小高度
        for i in range(n+1):
            if(i==0):
                dp[i]=0
            w=0
            h=0
            // 将前i - 1本书从第i - 1本开始放在与i同一层，直到这一层摆满或者所有的书都摆好
            for j in range(i-1,-1,-1):
                w+=books[j][0]
                h=max(h,books[j][1])
                if(w>shelf_width):
                    break
                dp[i]=min(dp[i],dp[j]+h)
        return dp[-1]

参考文献

​​[LeetCode] Leetcode 1105：填充书架（超详细的解法！！！）​​​​动态规划 Python3​​​​【LeetCode】1105. Filling Bookcase Shelves(填充书架)​​