先给出一个我自己写的方案，TLE
1 class Solution: 2 def __init__(self): 3 self.memo = dict() 4 self.memo[1] = True 5 6 def isUgly(self,num): 7 if num <= 0: 8 return False 9 if num == 1: 10 return True 11 if num % 2 == 0: 12 p = num // 2 13 if p in self.memo: 14 return self.memo[p] 15 else: 16 return self.isUgly(p) 17 if num % 3 == 0: 18 p = num // 3 19 if p in self.memo: 20 return self.memo[p] 21 else: 22 return self.isUgly(p) 23 if num % 5 == 0: 24 p = num // 5 25 if p in self.memo: 26 return self.memo[p] 27 else: 28 return self.isUgly(p) 29 return False 30 31 def nthUglyNumber(self, n: int) -> int: 32 i,j = 1,2 33 while i != n: 34 self.memo[j] = self.isUgly(j) 35 if self.memo[j] == True: 36 i += 1 37 j += 1 38 return j - 1
再给出一个AC的方案，使用动态规划：
1 class Solution: 2 def nthUglyNumber(self, n: int) -> int: 3 dp = [1]*n 4 p2, p3, p5 = 0, 0, 0 5 for i in range(1, n): 6 cur = min(2*dp[p2], 3*dp[p3], 5*dp[p5]) 7 if cur == 2*dp[p2]: 8 p2 += 1 9 if cur == 3*dp[p3]: 10 p3 += 1 11 if cur == 5*dp[p5]: 12 p5 += 1 13 dp[i] = cur 14 return dp[-1]