DNF 的一个变体如下:
def dutch_flag_partition(pivot_index , A):
pivot = A[pivot_index]
# First pass: group elements smaller than pivot.
for i in range(len(A)):
# Look for a smaller element.
for j in range(i + 1, len(A)):
if A[j] < pivot:
A[i], A[j] = A[j], A[i]
break
# Second pass: group elements larger than pivot.
for i in reversed(range(len(A))):
if A[i] < pivot:
break
# Look for a larger element. Stop when we reach an element less than
# pivot , since first pass has moved them to the start of A.
for j in reversed(range(i)):
if A[j] > pivot:
A[i], A[j] = A[j], A[i]
break
额外的空间复杂度为 O(1)。那是因为交换不依赖于输入长度吗?时间复杂度,给定为 O(N^2),是因为嵌套循环吗?谢谢