丑数最快解法【python】

题目描述

把只包含质因子2、3和5的数称作丑数（Ugly Number）。例如6、8都是丑数，但14不是，因为它包含质因子7。 习惯上我们把1当做是第一个丑数。求按从小到大的顺序的第N个丑数。

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暴力算法

判断丑数的函数：

def isUglyNum(self, num):
     while num % 2 == 0:
         num /= 2
     while num % 3  == 0:
         num /= 3
     while num % 5 == 0:
         num /= 5
     if num == 1:
         return 1
     return 0

如果采用暴力的解法：

class Solution:
    def isUglyNum(self, num):
        while num % 2 == 0:
            num /= 2
        while num % 3  == 0:
            num /= 3
        while num % 5 == 0:
            num /= 5
        if num == 1:
            return 1
        return 0
    def GetUglyNumber_Solution(self, index):
        # write code here
        if index < 1:
            return 0
        if index == 1:
            return 1
        result = 1
        count = 1
        while True:
            result += 1
            if self.isUglyNum(result):
                count += 1
            if count == index:
                return result
print(Solution().GetUglyNumber_Solution(1500))

求1500个丑数肯定会超时

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改进算法

使用T2 T3 T5表示遍历的时候乘以2 3 5，需要遍历之前的每一个数字直到求出M2 M3 M5

class Solution:
    def GetUglyNumber_Solution(self, index):
        # write code here
        if index < 1:
            return 0
        if index == 1:
            return 1
        uglyNumberList = [1]
        T2 , T3, T5 = 0, 0, 0
        for i in range(1, index):
            if uglyNumberList[T2] * 2 <= uglyNumberList[i - 1]:
                T2 += 1
            if uglyNumberList[T3] * 3 <= uglyNumberList[i - 1]:
                T3 += 1
            if uglyNumberList[T5] * 5 <= uglyNumberList[i - 1]:
                T5 += 1
            uglyNumber = min(uglyNumberList[T2] * 2, uglyNumberList[T3] * 3, uglyNumberList[T5] * 5) #M2 M3 M5
            uglyNumberList.append(uglyNumber)
        return uglyNumberList[index - 1]
print(Solution().GetUglyNumber_Solution(1500))

输出：859963392