1,2,3,5,6,7,8 如何合并成1-3,5-8

rubyish

复杂的 slice_before
in doc Enumerable

1. a = [0, 2, 3, 4, 6, 7, 9]
   
2. prev = a[0]
   
3. puts a.slice_before { |e|
   
4.   prev, prev2 = e, prev
   
5.   prev2 + 1 != e
   
6. }.map { |es|
   
7.   es.length <= 2 ? es.join(",") : "#{es.first}-#{es.last}"
   
8. }.join(",")

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ddd010

1. #!/usr/bin/ruby -w
   
2. 
   
3. bR=[]
   
4. 
   
5. aR=DATA.gets.chomp.split(/,/)
   
6. 
   
7. aR.each_index { |x|
   
8. 
   
9.   j=1
   
10.   while (aR[x+j].to_i-aR[x+j-1].to_i==1) #(aR[x].to_i..aR[x+j].to_i).include?(aR[x].to_i)
    
11.     j+=1
    
12.   end
    
13.   if j>1
    
14.     bR<<[aR[x], aR[x+j-1]]
    
15.     #aR.delete_if { |n| n.to_i<aR[x+j].to_i && n.to_i>aR[x].to_i }
    
16.     (j-1).times { |i|
    
17.       aR.delete_at(x+i-1)
    
18.     }
    
19.   end
    
20. }
    
21. 
    
22. bR.each { |x|
    
23.   print x, ?\n
    
24. }
    
25. 
    
26. #puts aR
    
27. __END__
    
28. 1,2,3,5,6,7,8,12,13,14,18,19,20,21,22,23,24

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结果：

1. ["1", "3"]
   
2. ["5", "8"]
   
3. ["12", "14"]
   
4. ["18", "23"]

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那个帮我看看，怎么回事。

rubyish

1. 
   
2. (j - 1).times { |i| aR.delete_at(x + i - 1) }
   

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first time:
x = 0, j = 3
first i = 0
0 + 0 - 1 = -1 => you delete aR[-1] = "24"
try~

1. 
   
2. (j - 1).times { |i| aR.delete_at(0) }
   

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ddd010

rubyish 发表于 2014-01-20 01:03
first time:
x = 0, j = 3
first i = 0

谢谢，原来如此。