红宝石递归算法问题

Question

在算法上工作：

Given a rows x cols screen and a sentence represented by a list of non-empty words, find how many times the given sentence can be fitted on the screen.
Note:
A word cannot be split into two lines.
The order of words in the sentence must remain unchanged.
Two consecutive words in a line must be separated by a single space.
Total words in the sentence won't exceed 100.
Length of each word is greater than 0 and won't exceed 10.
1 ≤ rows, cols ≤ 20,000.
Example 1:
Input:
rows = 2, cols = 8, sentence = ["hello", "world"]
Output: 
1
Explanation:
hello---
world---
The character '-' signifies an empty space on the screen.
Example 2:
Input:
rows = 3, cols = 6, sentence = ["a", "bcd", "e"]
Output: 
2
Explanation:
a-bcd- 
e-a---
bcd-e-
The character '-' signifies an empty space on the screen.
Example 3:
Input:
rows = 4, cols = 5, sentence = ["I", "had", "apple", "pie"]
Output: 
1
Explanation:
I-had
apple
pie-I
had--
The character '-' signifies an empty space on the screen.

这是我的代码：

def words_typing(sentence, rows, cols)
   count_words(sentence, rows, cols, cols, 0, 0)
end
def count_words(sentence, rows, cols, remaining_space, row_num, word_idx)
    return 0 if row_num == rows #keep going until out of rows, ends the recursion
    word_idx = 0 if word_idx == sentence.length  #reset the word back to the first
    if remaining_space >= sentence[word_idx].length
        if remaining_space == sentence[word_idx].length
            return 1 + count_words(sentence, rows, cols, remaining_space - sentence[word_idx].length, row_num, word_idx + 1 )
        else #greater than 1
            return 1 + count_words(sentence, rows, cols, remaining_space - sentence[word_idx].length - 1, row_num, word_idx + 1 )
        end
    else #move to a new line, reset remaining space
        return count_words(sentence, rows, cols, cols, row_num+1, word_idx)
    end 
end

代码工作如下。word_idx是句子数组中单词的索引。剩余空间最初是列的数量。只要有足够的空间可以放下一个单词，我就会返回1 在同一行上使用下一个单词和剩余空间的函数调用。如果剩余的空间> = 1 单词长度，那么我将在两个连续单词之间考虑一个空间（这就是为什么我有额外的条件）。

如果Word_idx的长度比句子数组更长，则将其重置为零。递归功能将继续进行，直到Row_num现在大于问题中提供给我们的行数。

但是，此代码不起作用。我的输出通常大于正确的答案，但是从概念上讲，我似乎都可以。有人看到我的方法有问题吗？

Answer 1

这是因为您在计数单词而不是句子。

def words_typing(sentence, rows, cols)
   count_words(sentence, rows, cols, cols, 0, 0, 0)
end
def count_words(sentence, rows, cols, remaining_space, row_num, word_idx, number_of_sentences)
    nos = number_of_sentences
    return nos if row_num == rows #keep going until out of rows, ends the recursion
    if word_idx == sentence.length  #reset the word back to the first
    word_idx = 0 
    nos = number_of_sentences+1
    end
    if remaining_space >= sentence[word_idx].length
        if remaining_space == sentence[word_idx].length
            return count_words(sentence, rows, cols, remaining_space - sentence[word_idx].length, row_num, word_idx + 1, nos )
        else #greater than 1
            return count_words(sentence, rows, cols, remaining_space - sentence[word_idx].length - 1, row_num, word_idx + 1 , nos)
        end
    else #move to a new line, reset remaining space
        return count_words(sentence, rows, cols, cols, row_num+1, word_idx, nos)
    end 
end

rows = 3
 cols = 6
 sentence = ["a", "bcd", "e"]
words_typing(sentence, rows, cols)
rows = 4; cols = 5; sentence = ["I", "had", "apple", "pie"]
words_typing(sentence, rows, cols)

我介绍了一个新的变量/参数（最后一个），该变量/参数（最后一个）（在开始0）。当word_idx == sentence.length时，这意味着在其余空间中安装了新句子，因此nos = number_of_sentences+1。
最后，我们返回编号（句子数）。

Answer 2

当您确定问题时，我想提出另一种编写方法的方法。

def sentences_per_page(rows, cols, sentence)
  nbr_sentences = 0
  last_word_index = sentence.size-1
  loopy = sentence.each_with_index.cycle
  word, idx = loopy.next
  rows.times do
    cols_left = cols
    while cols_left >= word.size
      cols_left -= (word.size + 1)
      nbr_sentences += 1 if idx == last_word_index
      word, idx = loopy.next
    end
  end
  nbr_sentences
end
rows = 4
cols = 5
sentence = ["I", "had", "apple", "pie"]
puts                    "    rows      sentences"
(1..12).each { |n| puts "     %d           %d" %
  [n, sentences_per_page(n, cols, sentence)] }
rows      sentences
  1           0
  2           0
  3           1
  4           1
  5           1
  6           2
  7           2
  8           2
  9           3
 10           3
 11           3
 12           4

我已经使用了方法数组＃周期。对于上述定义的sentence，

loopy = sentence.each_with_index.cycle
  #=> #<Enumerator: #<Enumerator: ["I", "had", "apple", "pie"]:each_with_index>:cycle> 
loopy.first 10
  #=> [["I", 0], ["had", 1], ["apple", 2], ["pie", 3],
  #    ["I", 0], ["had", 1], ["apple", 2], ["pie", 3],
  #    ["I", 0], ["had", 1]]