Wins the Offer (thirty-three): Number of ugly

Wins the Offer (thirty-three): Number of ugly

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First, the primer

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Second, the title

It contains only the number of prime factors 2, 3 and 5 is referred to as the number of ugly (Ugly Number). E.g. 6,8 are several ugly, but not 14, because it contains seven prime factors. Traditionally we have 1 as the first ugly number. Seeking ascending through N ugly large number sequence.

1, ideas

M is a number called number n of another factor, refers m divisible by n, i.e. n% m == 0. The definition of the ugly, ugly numbers 2, 3 and 5 can only be divisible. The definition of the ugly, ugly ugly number should be multiplied by another number 2, 3, or 5 (except 1). Therefore, we can create an array, which number is the number of ugly row good sequence, each number is the number of ugly ugly front multiplied by 2, 3 or 5 get.

The key problem with this idea is how to ensure that the number of ugly inside the array is sorted. For multiplied by 2, there is definitely a certain number of ugly T2, each of a number of ugly ahead of its results will be multiplied by 2 to get less than the maximum number has been ugly, ugly each number after it is multiplied by the multiplier 2 with the results obtained will be too large. We only need to remember the position of the lower number of ugly, but every time a new number is generated when the ugly, to update this T2. Multiplied by 3 and 5, the same also exists T3 and T5.

2, programming

python

Code implementation:

# -*- coding:utf-8 -*-
class Solution:
    def GetUglyNumber_Solution(self, index):
        # write code here
        if index < 7:
            return index
        res = [1, 2, 3, 4, 5, 6]
        # res[t2]=4 4*2=8 大于res里面4*1的值  res[t3]=3 3*3=9 大于res里面3*2的值 res[t5]=2 5*2=10 大于res里面5*1的值 
        t2, t3, t5 = 3, 2, 1
        for i in range(6, index):
            res.append(min(res[t2] * 2, res[t3] * 3, res[t5] * 5))
            while res[t2] * 2 <= res[i]:
                t2 += 1
            while res[t3] * 3 <= res[i]:
                t3 += 1
            while res[t5] * 5 <= res[i]:
                t5 += 1
        return res[index - 1]

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