[Lintcode]4. Ugly Number II/[Leetcode]264. Ugly Number II

本文链接：https://blog.youkuaiyun.com/siriusli1981/article/details/87563254

本文探讨了如何高效地找到第N个丑数，即仅包含2、3、5为质因数的正整数。介绍了两种算法实现，一种是直接判断每个数是否为丑数，但效率较低；另一种是使用动态规划思想，通过维护三个指针来快速生成丑数序列，最终实现O(n)的时间复杂度。

4. Ugly Number II / 264. Ugly Number II

本题难度: Medium
Topic: Data Structure

Description

Ugly number is a number that only have factors 2, 3 and 5.

Design an algorithm to find the nth ugly number. The first 10 ugly numbers are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12…

Example
Example 1:

Input: 9
Output: 10
Example 2:

Input: 1
Output: 1
Challenge
O(n log n) or O(n) time.

Notice
Note that 1 is typically treated as an ugly number.

我的代码超时。因为每个数都要判断一次

    def nthUglyNumber(self, n: 'int') -> 'int':
        cur = 1
        while(n-1):
            cur += 1
            if self.isUgly(cur):
                n -= 1       
        return cur
    def isUgly(self, n):
        ugly = [2,3,5]
        for i in ugly:
            while(n % i == 0):
                n //= i
        return n == 1

别人的代码

class Solution:
    def nthUglyNumber(self, n: 'int') -> 'int':
        res = [1]
        i2 = i3 = i5 = 0
        while(n>1):
            u2, u3, u5 = 2*res[i2], 3*res[i3], 5*res[i5]
            uCurr = min(u2,u3,u5)
            if uCurr == u2:
                i2 += 1
            if uCurr == u3:
                i3 += 1
            if uCurr == u5:
                i5 += 1
            res.append(uCurr)
            n -= 1
        return res[-1]