Question

Q-我有一个大小为apples的数组A，并且必须创建另一个数组S，该数组将按排序顺序包含苹果的索引，因为我们不能直接访问或触摸A，而只能是一个函数is_large（A，i ，j）函数可以访问它。如果A [i]> A [j]，则返回-1，如果A [i]

def is_large_apples(apple_size, i, j):
    """ Takes two indices and tells
    which one of them is larger or smaller """

    if apple_size[i] > apple_size[j]:
        return 1
    elif apple_size[i] < apple_size[j]:
        return -1

def mergesort_apples(s, l, r):
    """ This function takes indexed list and
     makes recursive calls to sort the array """
    if l < r:
        mid = (l+r)//2
        mergesort_apples(s, l, mid)
        mergesort_apples(s, mid+1, r)
        merge_apples(s, l, mid, r)


def merge_apples(s, l, mid, r):
    """ This function takes the list and
    the indices to merge them into the final array"""
    nl = mid - l + 1
    nr = r - mid
    left, right = [], []
    left = s[l:mid+1:1]
    right = s[mid+1:r+1:1]
    i, j, k = 0, 0, l;
    while i < nl and j < nr:
        print(s)
        if is_large_apples(apple_size, i, j) == -1:
            s[k] = left[i]
            i += 1
        else:
            s[k] = right[j]
            j += 1
        k += 1

    while i < nl:
        s[k] = left[i]
        k += 1
        i += 1
    while j < nr:
        s[k] = right[j]
        k += 1
        j += 1

apple_size = [5, 7, 1,44,2,33] # Given list of sizes.
s = [x for x in range(0,len(apple_size))] # Original list of indices.
mergesort_apples(s, 0, len(s)-1)
print(s)

Answer 1

if is_large_apples(apple_size, left[i], right[j]) == -1:

因为您不想检查i和j的位置，而是要检查left[i]的位置和right[j]的位置。

使用合并排序原则

1 个答案: