264. Ugly Number II

Description

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, and n does not exceed 1690.

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

Solution

Three-pointer, time O(n), space O(n)

The ugly-number sequence is 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, …
because every number can only be divided by 2, 3, 5, one way to look at the sequence is to split the sequence to three groups 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, …
We can find that every subsequence is the ugly-sequence itself (1, 2, 3, 4, 5, …) multiply 2, 3, 5.

Then we use similar merge method as merge sort, to get every ugly number from the three subsequence.

Every step we choose the smallest one, and move one step after,including nums with same value. 注意三个factor有可能相等的！被选中的时候，相同的factor都需要向前移动。

class Solution {
    public int nthUglyNumber(int n) {
        if (n < 1) {
            return -1;
        }
        
        int[] ugly = new int[n];
        ugly[0] = 1;
        int index2 = 0, index3 = 0, index5 = 0;     // use three pointers
        int factor2 = 2, factor3 = 3, factor5 = 5;
        
        for (int i = 1; i < n; ++i) {
            int min = Math.min(factor2, Math.min(factor3, factor5));
            ugly[i] = min;
            // move one step after for min,including nums with same value
            if (factor2 == min) {
                factor2 = 2 * ugly[++index2];
            }
            if (factor3 == min) {
                factor3 = 3 * ugly[++index3];
            }
            if (factor5 == min) {
                factor5 = 5 * ugly[++index5];
            }
        }
        
        return ugly[n - 1];
    }
}