对齐字符串红宝石

Question

需要做的理由对于给定的字符串与所述给定长度，

ss = "There is an example to make Justify the text us" 

Total length = 50 

需要的输出：

ss = "There is an example to make Justify the text us" 

这里串的长度与50字符调节之后加入间距第一个字符“There”，最后一个字符“us”，“text”。

都试过了，

qw = "" 
string.split(" ").each_with_index do |x,i| 
qw = qw + " " + x 

end 

，但不能能够增加空间

Answer 1

假设你想先添加空间最后一个字，然后在第一，然后到倒数第二等：

def justify(str, length) 
    words = str.split(/\s+/) 
    cnt = words.size - 1 
    return str if cnt < 1 
    return str if str.size >= length 

    diff = length - str.size 
    base = diff/cnt + 1 
    rest = diff % cnt 

    cnt.times.each.with_index do |word, i| 
    r = [i + 1, cnt - i - 1].min * 2 <= rest && rest > 0 ? 1 : 0 
    rest -= r 
    words[i] << " " * (base + r) 
    end 
    words.join 
end 

justify("There is an example to make Justify the text us", 50) 
#=> "There is an example to make Justify the text us" 

justify("One more example", 50) 
#=> "One     more     example" 

justify("Or this", 50) 
#=> "Or           this" 

justify("Example", 50) 
#=> "Example" 

justify("There is an example to make Justify the very long text us", 50) 
#=> "There is an example to make Justify the very long text us" 

Answer 2

您的示例代码是不工作的原因是split消耗的分隔符。

'a b c'.split.to_a #=> ['a', 'b', 'c'] 

当您手动重装你的字符串，你加入，你消耗的界定空间，导致相同的字符串一个空格（）。

正如评论所建议的那样，您尚未向我们提供足够的信息来说明您想如何处理此问题。如果只有一个字会发生什么？如果我们有更多的字符而不是线条应该是什么？你想如何称重空间？你需要考虑文本的渲染宽度吗？这很快成为一个更复杂的项目。

但是，请注意，大多数视图（无论是移动设备，网页还是外壳）都有方法来验证您可以利用的文本的效果。

这一切都表示，这里是一种工作示例中忽略的边缘情况：

input = "There is an example to make Justify the text us" 
words = input.split 
spaces_needed = 50 - input.scan(/[^ ]/).length 
spaces_per_word = Float(spaces_needed)/(words.count - 1) 

words.each_with_index.map do |word, i| 
    word += ' ' if (spaces_per_word * (i + 1)) % 1 == 0 
    word 
end.join(' ') 

Answer 3

试试这个，

str = 'There is an example to make Justify the text us' 

words = str.split 
width, remainder = (50 - words.map(&:length).inject(:+)).divmod(words.length - 1) 
width, remainder = 1, 0 if width.zero? # take care of long lines 
words.take(words.length - 1).each { |word| width.times { word << 32 }} 
words.take(words.length - 1).shuffle.take(remainder).each { |word| word << 32 } 
p words.join 

这是如何工作的？

• 分割字符串成字
• 计算的空间宽度和余k
• 附加空间宽度到所有的话，但最后一个
• 追加一个空间的k话

Answer 4

随机样本

这里有一个合理可读的方式来做到这一点：

def padString (str,width) 
    words = str.split.map{|s|s+" "}  # words single spaced 
    padding = width-words.join.length+1 # how much needs adding? 
    locations = words.size - 1   # places space can be added 
    return str if padding<0 or words==[] # abort if string too long or empty 
    padding.times { 
     words[rand(locations)] += " " # append space to random location 
    } 
    words.join.strip!     # reconstitute the string 
end 

它随机添加填充，所以你得到不同的线通过每个时间：

padString("There is an example to make Justify the text us",50) 
=> "There is an example to make Justify the text us" 

padString("There is an example to make Justify the text us",50) 
=> "There is an example to make Justify the text us" 

padString("hey you",50) 
=> "hey           you" 

padString("hey",50) 
=> "hey"