Question

我有一个问题：

我们有一个整数数组（大小为n）require_command!' from /home/frank/.rbenv/versions/2.2.4/lib/ruby/gems/2.2.0/gems/railties-4.2.5.1/lib/rails/commands/commands_tasks.rb:130:in

我们必须找到generate' from /home/frank/.rbenv/versions/2.2.4/lib/ruby/gems/2.2.0/gems/railties-4.2.5.1/lib/rails/commands/commands_tasks.rb:39:in f.e <top (required)>' from bin/rails:4:in的最小公共部分[x_1,x_2,...,x_n]（最长的正确部分是[x_i,x_i+1,...,x_j]）

我无法弄清楚如何将时间减少到(x_i,x_i+1,...,x_j) >= j-i+1以下（检查每一段）我的第二个想法是：每个值[2,2,1,4,3,3,1] -> 3等于＆＃34;长度范围＆＃34;这个部分可能有，如果数量较少那么我们改变＆＃34;长度范围＆＃34; （同时计算这个细分的时间长度），但如果我们得到更多的数字，我也不知道该怎么做

（我会非常感谢任何帮助 - 可能还有另一个更简单的解决方案，但我不会看到它）

Answer 1

Caterpillar method, O(n), assuming all integers will be positive:

Put a left and right pointer on the first number, and your minimum = this number. Length = 1
Advance the right pointer by one, check for the condition (you know if there's a new minimum very easily, and you know the length += 1)
If you cannot advance the right pointer without breaking the condition, advance the left pointer and update your values
If you cannot advance the left pointer either, you have found a section satisfying the condition. Note its length, and start a new section in the same way, on the next index.

Example for you:

[2,2,1,4,3,3,1]
[2] <- The first 2
[2,2] <- Moved the right
[2] <- Moved the left, both R and L are pointing at the second 2
<-Can't move either, note max length and indices from previous run.
[1] <- Starting again
<-Can't move either
[4] <- Starting again
[4,3] <- Right
[4,3,3] <- Right (New max length and indices noted)
[3,3] <- Left. Doesn't improve the max
[3] <- Left again.
<- Can't move either, note max and indices, in this case overwrite length
[1] <-Start again
<- end of loop, check global max length, output

This is O(n) because each pointer (L and R) only moves over any given index once.

I think this should be enough for you to write the algorithm in any language you choose. Look out for the end-of-loop condition, because you have to check against the global max length if you run out of array.

算法 - 最小大于大小

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