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.

 

Approach #1: C++. generate all ugly numbers then sort.

class Solution {
public:
    int nthUglyNumber(int n) {
        vector<int> temp;
        for (long i = 1; i < INT_MAX; i = i*2) {
            for (long j = i; j < INT_MAX; j = j*3) {
                for (long k = j; k < INT_MAX; k = k*5) {
                    temp.push_back(k);
                }
            }
        }
        sort(temp.begin(), temp.end());
        return temp[n-1];
    }
};

　　

Approach #2: Java. [order generate]

class Solution {
    public int nthUglyNumber(int n) {
        List<Integer> nums = new ArrayList<Integer>();
        nums.add(1);
        int i1 = 0;
        int i2 = 0;
        int i3 = 0;
        while (nums.size() < n) {     
            int ant = Math.min(nums.get(i1) * 2, Math.min(nums.get(i2) * 3, nums.get(i3) * 5));
            nums.add(ant);
            if (nums.get(i1) * 2 == ant) i1++;
            if (nums.get(i2) * 3 == ant) i2++;
            if (nums.get(i3) * 5 == ant) i3++;
        }
        
        return nums.get(n-1);
    }
}

　　

Approach #3: Python.

class Solution(object):
    def nthUglyNumber(self, n):
        """
        :type n: int
        :rtype: int
        """
        ugly = [1]
        i2 = i3 = i5 = 0
        
        while len(ugly) < n:
            while ugly[i2]*2 <= ugly[-1]: i2 += 1
            while ugly[i3]*3 <= ugly[-1]: i3 += 1
            while ugly[i5]*5 <= ugly[-1]: i5 += 1
            ugly.append(min(ugly[i2]*2, ugly[i3]*3, ugly[i5]*5))
        
        return ugly[-1]