Java 将煎饼排序的循环实现转换为递归实现
我正在学习递归,想把我的循环转换成递归函数?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Java 将煎饼排序的循环实现转换为递归实现,java,recursion,Java,Recursion,我正在学习递归,想把我的循环转换成递归函数?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.
/**
* 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,我不想写你的程序,如果这是私人的,我也不想破坏你的乐趣,所以我在Ruby中实现了一个有大量注释的递归解决方案。它基本上归结为进行必要的翻转,将最大元素移动到数组的末尾,然后将相同的逻辑应用于通过排除最大值创建的子数组。递归调用返回已排序的子数组,因此只需追加最大值并将结果返回行。递归的基本情况是,当您开始处理单个元素时,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]
函数查找数组中max元素的索引。
#如果是平局,则返回最低索引
def max_索引(a)
a、 每个| index.inject{| memo,i | a[i]>a[memo]?i:memo}
结束
def煎饼分类(a)
#对包含0或1个元素的数组进行排序(一般),
#把它还给我就行了。这是递归的基本情况。
如果a.length<2,则返回a
#找到最大值的位置,将其表示为第n个元素
n=最大指数(a)+1
#翻转前n个元素的堆栈(最大值为最后一个)
#要将最大值放在前面,请将其与
#数组的其余部分,然后翻转整个结果。这
#会把max放在最后。然而,不要为所有的事情烦恼
#如果max已经到了终点的话。
a=(a.take(n)。反转+a.drop(n))。如果nif(i),您缺少return
语句