leetcode264

先给出一个我自己写的方案，TLE

class Solution:
    def __init__(self):
        self.memo = dict()
        self.memo[1] = True

    def isUgly(self,num):
        if num <= 0:
            return False
        if num == 1:
            return True
        if num % 2 == 0:
            p = num // 2
            if p in self.memo:
                return self.memo[p]
            else:
                return self.isUgly(p)
        if num % 3 == 0:
            p = num // 3
            if p in self.memo:
                return self.memo[p]
            else:
                return self.isUgly(p)
        if num % 5 == 0:
            p = num // 5
            if p in self.memo:
                return self.memo[p]
            else:
                return self.isUgly(p)
        return False

    def nthUglyNumber(self, n: int) -> int:
        i,j = 1,2
        while i != n:
            self.memo[j] = self.isUgly(j)
            if self.memo[j] == True:
                i += 1
            j += 1
        return j - 1

再给出一个AC的方案，使用动态规划：

class Solution:
    def nthUglyNumber(self, n: int) -> int:
        dp = [1]*n
        p2, p3, p5 = 0, 0, 0
        for i in range(1, n):
            cur = min(2*dp[p2], 3*dp[p3], 5*dp[p5])
            if cur == 2*dp[p2]:
                p2 += 1
            if cur == 3*dp[p3]:
                p3 += 1
            if cur == 5*dp[p5]:
                p5 += 1
            dp[i] = cur
        return dp[-1]