Question

def binary_search(num, test_array)
  counter = 0
  low = 0
  high = test_array.length

  while (low <= high) do
    i = ((low + high) / 2).floor;
    if num == test_array[i]
      return i
      break
    elsif num < test_array[i]
      high = i
    elsif num > test_array[i]
      low = i
    end
    counter += 1
  end
end

test_array = [13, 19, 24, 29, 32, 37, 43]
# binary_search(35, test array)?
# binary_search(11, test array)?

在此代码中，我如何确保它将返回＆＃34; nil＆＃34;如果是binary_search（35）或binary_search（11）？

Answer 1

正如@YuHao指出的那样，当你的元素找不到时，它是一个无限循环：low = high = 0（在低端），这使循环无限期地运行。

def binary_search(num, test_array)
  counter = 0
  low = 0
  high = test_array.length

  while (low <= high) do
    i = ((low + high) / 2).floor;
    if num == test_array[i]
      return i
    elsif num < test_array[i]
      high = i
    elsif num > test_array[i]
      low = i
    end
    counter += 1

    break if counter > test_array.length / 2 # could probably be more efficient
  end

  nil
end

test_array = [13, 19, 24, 29, 32, 37, 43]

binary_search(11, test_array) # => nil
binary_search(35, test_array) # => nil

Answer 2

正如余浩所说，当什么都没找到时，它会进入一个无限循环。以下是检查无限循环情况并返回nil的示例。

def binary_search(num, test_array)
  counter = 0
  low = 0
  high = test_array.length

  while (low <= high) do
    i = ((low + high) / 2).floor;
    if num == test_array[i]
      return i
      break
    elsif num < test_array[i]
      high = i
    elsif num > test_array[i]
      low = i
    end
    counter += 1

    # check for value not found
    if (high - low) <= 1
      return nil
    end

  end
end

put＆＃34; nil＆＃34;如果它不在范围内

2 个答案: