targetofferbyPython 稀疏矩阵转置、丑数优化 about algorithmsbypython HOT 1 OPEN

丑数哪一题 没必要加循环，可以简单判断最小的丑数包含几个2,3,5
def solution(index: int) -> int:
'''
1 -> 1
2 -> 2 12
3 -> 3 13
4 -> 4 22
5 -> 5 15
6 -> 6 23
7 -> 8 222
8 -> 9 33
9 -> 10 25
10 -> 12 223
11 -> 15 35
直接判断这个数里面分别包含几个2,3,5
'''
if index < 0:
return
ants = [1] * index
index2, index3, index5 = 0, 0, 0
for i in range(1,index):
ants[i] = min(ants[index2] * 2, ants[index3] * 3, ants[index5] * 5)
if ants[index2] * 2 == ants[i]:
index2 += 1
if ants[index3] * 3 == ants[i]:
index3 += 1
if ants[index5] * 5 == ants[i]:
index5 += 1
return ants[-1]

print(solution(11))

from algorithmsbypython.