[英]Converting loop implementation of Pancake Sort to a recursive implementation
我正在学习递归,并想将我的循环转换为递归函数? 此代码的正确答案应该是什么(假设我已经编写了flip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 );
}
如果这是家庭作业,我不想编写您的程序,或者如果这是个人的,我不想破坏您的乐趣,因此我在Ruby中实现了一个受到广泛评论的递归解决方案。 它基本上归结为进行必要的翻转,以将最大元素移到数组的末尾,然后将相同的逻辑应用于通过排除最大值而创建的子数组。 递归调用返回排序后的子数组,因此只需追加max并将结果传递回该行即可。 递归的基本情况是当您深入到单个元素时,只需将其返回即可。
# 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]
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