lintcode 丑数 python

暴力如下：

class Solution:
    """
    @param n: An integer
    @return: return a  integer as description.
    """
    def istrue(self,x):
        a=[i for i in range(3)]
        a[0],a[1],a[2]=2,3,5
        flag=1
        while x!=1:
            flag1=0
            for i in range(2,-1,-1):
                if x%a[i]==0:
                    x/=a[i]
                    flag1=1
                    break
            if flag1==0:
                flag=0
                break
        return flag
    def nthUglyNumber(self, n):
        # write your code here
        index=0
        answer=0
        for i in range(1,100000*n):
            if self.istrue(i)==1:
                index+=1
                answer=i
            if index==n:
                break
        return answer

这样是肯定过不了的，必然T

用heapq模板中的heappush heappop

from heapq import heappop,heappush
class Solution:
    def nthUglyNumber(self, n):
        heap=[1]
        x=[2,3,5]
        y=[]
        for _ in range(n):
            num=heappop(heap)
            y=[i*num for i in x]
            for i in y:
                if i not in heap:
                    heappush(heap,i)
        return num