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ugly-number-ii.py 2.05 KB
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Jack 提交于 8年前 . add some solutions
# Time: O(n)
# Space: O(1)
#
# Write a program to find the n-th ugly number.
#
# Ugly numbers are positive numbers whose prime factors
# only include 2, 3, 5. For example,
# 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the
# first 10 ugly numbers.
#
# Note that 1 is typically treated as an ugly number.
#
# Hint:
#
# The naive approach is to call isUgly for every number
# until you reach the nth one. Most numbers are not ugly.
# Try to focus your effort on generating only the ugly ones.
#
import heapq
class Solution:
# @param {integer} n
# @return {integer}
def nthUglyNumber(self, n):
ugly_number = 0
heap = []
heapq.heappush(heap, 1)
for _ in xrange(n):
ugly_number = heapq.heappop(heap)
if ugly_number % 2 == 0:
heapq.heappush(heap, ugly_number * 2)
elif ugly_number % 3 == 0:
heapq.heappush(heap, ugly_number * 2)
heapq.heappush(heap, ugly_number * 3)
else:
heapq.heappush(heap, ugly_number * 2)
heapq.heappush(heap, ugly_number * 3)
heapq.heappush(heap, ugly_number * 5)
return ugly_number
def nthUglyNumber2(self, n):
ugly = [1]
i2 = i3 = i5 = 0
while len(ugly) < n:
while ugly[i2] * 2 <= ugly[-1]: i2 += 1
while ugly[i3] * 3 <= ugly[-1]: i3 += 1
while ugly[i5] * 5 <= ugly[-1]: i5 += 1
ugly.append(min(ugly[i2] * 2, ugly[i3] * 3, ugly[i5] * 5))
return ugly[-1]
def nthUglyNumber3(self, n):
q2, q3, q5 = [2], [3], [5]
ugly = 1
for u in heapq.merge(q2, q3, q5):
if n == 1:
return ugly
if u > ugly:
ugly = u
n -= 1
q2 += 2 * u,
q3 += 3 * u,
q5 += 5 * u,
class Solution2:
ugly = sorted(2**a * 3**b * 5**c
for a in range(32) for b in range(20) for c in range(14))
def nthUglyNumber(self, n):
return self.ugly[n-1]
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