Question

我正在学习递归，并希望将我的循环转换为递归函数？这段代码的正确答案应该是什么（假设我已经编写了翻转方法来反转数组中的元素）？谢谢，

/**
 * Sorts an array of integers by repeatedly reversing 
 * subranges within the array. Prints the flip sequence. 
 */ 
public static void  sort( int[] array)
{   
    int size = array.length;
    if (!Ordered(array, size)){
        for(int i = size-1; i > 0; i--)
        {
            int j = findMax(array, 0, i );
            int flipPosition;

            if( j != i )
            {
                if( j != 0 ) {
                    flip( array, 0, j );
                    flipPosition = size-j;
                    System.out.print( flipPosition + " " );
                }

                flip( array, 0, i );
                flipPosition = size-i;
                System.out.print( flipPosition + " " );
            }   
        }
    }
    System.out.println( 0 );
}

Answer 1

如果这是家庭作业，我不想写你的程序，或者如果这是个人的话会毁了你的乐趣，所以我在Ruby中实现了一个评论很多的递归解决方案。它基本上归结为将最大元素移动到数组末尾所需的翻转，然后将相同的逻辑应用于通过排除max而创建的子数组。递归调用返回已排序的子数组，因此只需附加最大值并将结果传递回该行。递归的基本情况是当你到达单个元素时，只需返回它。

# Function to find the index of the max element in an array.
# In case of ties, returns the lowest index
def max_index(a)
  a.each_index.inject { |memo, i| a[i] > a[memo] ? i : memo }
end

def pancake_sort(a)
  # An array with 0 or 1 elements is sorted (trivially),
  # just return it.  This is the base case for the recursion.
  return a if a.length < 2

  # Find location of the max, express it as the n'th element
  n = max_index(a) + 1

  # Flip the stack of the first n elements (max is last)
  # to put the max at the front, concatenate it with the
  # rest of the array, then flip the entire result.  This
  # will put max at the end. However, don't bother with all
  # that flipping if max was already at the end.
  a = (a.take(n).reverse + a.drop(n)).reverse if n < a.length

  # Recursively apply the logic to the subarray that excludes
  # the last (max) element.  When you get back the sorted
  # subarray, tack the max back onto the end
  return pancake_sort(a.take(a.length - 1)) << a[-1]
end

# Create an array of 20 random numbers between 0 and 99    
ary = Array.new(20) { rand(100) }
# Display the result
p ary
# Display the result after sorting
p pancake_sort(ary)

# Sample output:
# [70, 19, 95, 47, 87, 49, 53, 8, 89, 33, 22, 85, 91, 87, 99, 56, 15, 27, 75, 70]
# [8, 15, 19, 22, 27, 33, 47, 49, 53, 56, 70, 70, 75, 85, 87, 87, 89, 91, 95, 99]

将Pancake Sort的循环实现转换为递归实现

1 个答案: