1105. Filling Bookcase Shelves解题报告

Description:

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Example:

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Link:

https://leetcode.com/problems/filling-bookcase-shelves/

解题方法：

DP:
dp[i] represents the minimum height of total shelves we build that ended by books[i].
For a given book with index i, assume this book is the last book in current level, then we can try to add other books with index < i into our current level, unless the total width of current level is bigger than the limit shelf_width. During this process, we will find out the minimum height of total shelves we've build, which is the dp[i] we are trying to figure out for current book.

dp[i]代表以当前这本书为本层书架最后一本书的时候，目前我们建的书架的最小高度。
当我们计算到书本i的时候，假设这本书是当前level最后一本书，那么我们可以尝试把之前的书本加入到本层书架来，只要这些书本的宽度不超过规定的宽度，在这个过程中。我们会找到以当前这本书为本层书架最后一本书为前提条件的时候，我们目前建立的书架最小的高度。然后把它更新到dp数组中。最后一个值就是我们的答案。

Time Complexity：

O（n^2)

完整代码：

class Solution {
public:
    int minHeightShelves(vector>& books, int shelf_width) {
        vector dp;
        int size = books.size();
        if(!size)
            return 0;
        dp.push_back(books[0][1]);
        for(int i = 1; i < size; i++) {
            int currWidth = books[i][0];
            int maxHeight = books[i][1];
            int minDp = maxHeight + dp.back();
            for(int j = dp.size() - 1; j >= 0; j--) {
                currWidth += books[j][0];
                if(currWidth > shelf_width)
                    break;
                maxHeight = max(maxHeight, books[j][1]);
                if(j == 0)
                    minDp = min(maxHeight, minDp);
                else
                    minDp = min(maxHeight + dp[j - 1], minDp);
            }
            dp.push_back(minDp);
        }
        return dp.back();
    }
};