Question

使用算法：

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

代码的工作原理如下。 word_idx是句子数组中单词的索引。剩余空间最初是列数。只要有足够的空间放下一个单词，我就会在同一行返回1 +函数调用，并且下一个单词和剩余的空间。如果剩余空间> = 1 +字长，那么我将考虑在两个连续单词之间留一个空格（这就是我有额外条件的原因）。

如果word_idx比句子数组长，它会重置为零。递归函数将继续运行，直到row_num现在大于问题中提供给我们的行数。

然而，这段代码并不起作用。我的输出通常大于正确的答案，但从概念上讲，所有这些似乎都适合我。有人看到我的做法有问题吗？

Answer 1

这是因为你在计算单词而不是句子。

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)

我引入了一个包含多个句子的新变量/参数（最后一个）（在开始0）。当word_idx == sentence.length表示新句子符合剩余空格时，nos = number_of_sentences+1 最后我们返回nos（句子数）。

Answer 2

当您发现问题时，我想建议另一种方法来编写该方法。

sentence

我使用了方法Array#cycle。对于上面定义的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]]，

testFunction

Ruby递归算法问题

2 个答案: