264. 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. 

Example:

Input: n = 10
Output: 12
Explanation: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.

Note:  

1. 1 is typically treated as an ugly number.
2. n does not exceed 1690.

Hint:

1. The naive approach is to call isUgly() for every number until you reach the n-th 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: L_1L​1​​, L_2L​2​​, and L_3L​3​​.
4. Assume you have U_kU​k​​, the k^{th}k​th​​ ugly number. Then U_{k+1}U​k+1​​ must be Min(L_1 * 2, L_2 * 3, L_3 * 5)Min(L​1​​∗2,L​2​​∗3,L​3​​∗5).

我信你个鬼，hint1我试了，果然超时。

分析

根据提示中的信息，我们知道丑陋序列可以拆分成3个子序列：

1. 1x2, 2x2, 3x2, 4x2, 5x2, ...
2. 1x3, 2x3, 3x3, 4x3, 5x3, ...
3. 1x5, 2x5, 3x5, 4x5, 5x5, ...

注意是1，2，3，4，5，6，8，9，10.。。。而不是1，2，3，4，5，6，7，8，9，，，，，，所以才能用存在数组里的数继续做

每次从三个列表中取出当前最小的那个加入序列，直到第n个为止。

public int nthUglyNumber(int n) {
        int[] nums = new int[n];
        int index2 = 0, index3 = 0, index5 = 0;
        nums[0] = 1;
        for(int i = 1; i < nums.length; i++){
            nums[i] = Math.min(nums[index2] * 2, Math.min(nums[index3] * 3, nums[index5] * 5));
            if(nums[i] == nums[index2] * 2)
                index2++;
            if(nums[i] == nums[index3] * 3)
                index3++;
            if(nums[i] == nums[index5] * 5)
                index5++;
        }
        return nums[n - 1];
    }

总结：

Ugly numbers can be created from2, 3, 5, then 

1. 1x2, 2x2, 3x2, 4x2, 5x2, ...
2. 1x3, 2x3, 3x3, 4x3, 5x3, ...
3. 1x5, 2x5, 3x5, 4x5, 5x5, ...

Since it was made of these 3 factors, we can use 3 variables to keep track the index of each factor. nums[index2/3/5] * 2/3/5