class Solution(object):
def nthUglyNumber(self, n):
"""
:type n: int
:rtype: int
"""
if n <= 5:
return n
dp = [0] * (n + 1)
l1 = l2 = l3 = 1
dp[1] = 1
dp[2] = 2
dp[3] = 3
dp[4] = 4
dp[5] = 5
for i in range(6, n + 1):
while dp[l1] * 2 <= dp[i - 1]:
l1 += 1
while dp[l2] * 3 <= dp[i - 1]:
l2 += 1
while dp[l3] * 5 <= dp[i - 1]:
l3 += 1
print l1, l2, l3
dp[i] = min(dp[l1] * 2, dp[l2] * 3, dp[l3] * 5)
# print dp
return dp[n]
if __name__ == '__main__':
# begin
s = Solution()
print s.nthUglyNumber(10)
Download Ugly Number II.pyLeetcode 264 Ugly Number II problem solution in python3 with explanation. This is the best place to expand your knowledge and get prepared for your next interview.
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