LeetCode----Ugly NumberII

Ugly Number II

Write a program to find the n-th ugly number.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.

Note that 1 is typically treated as an ugly number.

Hint:

1. The naive approach is to call isUgly for every number until you reach the nth one. Most numbers are not ugly. Try to focus your effort on generating only the ugly ones.
2. An ugly number must be multiplied by either 2, 3, or 5 from a smaller ugly number.
3. The key is how to maintain the order of the ugly numbers. Try a similar approach of merging from three sorted lists: L1, L2, and L3.
4. Assume you have Uk, the kth ugly number. Then Uk+1 must be Min(L1 * 2, L2 * 3, L3 * 5).

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

分析：

没看Hint之前，我以为可以用母函数来做，后来发现母函数之间的组合是加法，这里是乘法。之后又使用三层循环暴力组合2，3，5，结果一直超时。

代码1，使用最小堆：

class Solution(object):
    def nthUglyNumber(self, n):
        """
        :type n: int
        :rtype: int
        """
        from heapq import heappop, heappush
        h2, h3, h5 = [2], [3], [5]
        min_num = 1
        while n > 1:
            u2, u3, u5 = h2[0], h3[0], h5[0]
            min_num = min(u2, u3, u5)
            if min_num == u2:
                heappop(h2)
                heappush(h2, u2 * 2)
                heappush(h3, u2 * 3)
                heappush(h5, u2 * 5)
            if min_num == u3:
                heappop(h3)
                heappush(h3, u3 * 3)
                heappush(h5, u3 * 5)
            if min_num == u5:
                heappop(h5)
                heappush(h5, u5 * 5)
            n -= 1
        return min_num

使用了三个堆，使用堆会使时间消耗过大。

代码2：

class Solution(object):
    def nthUglyNumber(self, n):
        """
        :type n: int
        :rtype: int
        """
        uglylst = [1]
        i2, i3, i5 = 0, 0, 0  # represent the uglylst index
        while n > 1:
            u2, u3, u5 = uglylst[i2] * 2, uglylst[i3] * 3, uglylst[i5] * 5
            min_num = min(u2, u3, u5)
            if min_num == u2:
                i2 += 1
            if min_num == u3:
                i3 += 1
            if min_num == u5:
                i5 += 1
            uglylst.append(min_num)
            n -= 1
        return uglylst[-1]