Question:

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:

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.
An ugly number must be multiplied by either 2, 3, or 5 from a smaller ugly number.
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.
Assume you have Uk, the kth ugly number. Then Uk+1 must be Min(L1 2, L2 3, L3 * 5).

Thinking:

The main idea of this question is to maintain a min heap and get the minimum from the heap once. Because we can see that, the ugly numbers are as below:

(1) 1×2, 2×2, 3×2, 4×2, 5×2, …
(2) 1×3, 2×3, 3×3, 4×3, 5×3, …
(3) 1×5, 2×5, 3×5, 4×5, 5×5, …

Furthermore, we can record different factor’s index in the primes in order to make the algorithm more effienct.

Solution:

(LTE)

public int nthUglyNumber1(int n){
    List<Integer> res = new ArrayList<Integer>();
    PriorityQueue<Integer> heap = new PriorityQueue<Integer>();
    heap.add(1);
    int[] primes = {2, 3, 5};
    while (res.size() != n) {
        int num = heap.poll();
        if (res.size() == 0 || num != res.get(res.size()-1))
            res.add(num);
        for ( int j = 0; j < 3; j++){
            heap.add(num * primes[j]);
        }
    }

    return res.get(n-1);
}

(AC)

public int nthUglyNumber2(int n) {
    List<Integer> res = new ArrayList<Integer>();
    res.add(1);
    int[] index = new int[3];
    int[] list = new int[3];
    int[] primes = {2, 3, 5};
    list[0] = 1;
    list[1] = 1;
    list[2] = 1;

    for (int i = 1; i < n; i++){
        for (int j = 0; j < 3; j++){
            list[j] = res.get(index[j]) * primes[j];
        }
        res.add(Math.min(Math.min(list[0], list[1]), list[2]));
        for (int j = 0; j < 3; j++){
            if (list[j] == res.get(res.size()-1))
                index[j]++;
        }
    }

    return res.get(n-1);
}

Reference: https://leetcode.com/discuss/52716/o-n-java-solution