264. Ugly Number II

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.

My Solutions:

建立一个长度为n的数组nums。把2的倍数index2， 3的倍数index3，和5的倍数index5都设置为0。这些index代表由2/3/5乘以index之前，数字在nums中的index。必须要用nums里的数字和2/3/5相乘，这样产出的新数字只会被2/3/5整除。

nums的index = 0时数字是1，之后的每一位，找到2，3，5倍数后最小数。必须用num里已经有的数字乘以

0. nums=[1]

1. starts = [0,0,0] for numbers 2,3,5, so new_num = min(1*2,1*3,1*5) = 2, and now starts = [1,0,0], Numbers = [1,2].

2. starts = [1,0,0], so new_num = min(2*2,1*3,1*5) = 3, and now starts = [1,1,0], Numbers = [1,2,3].

3. starts = [1,1,0], so new_num = min(2*2,2*3,1*5) = 4, so now starts = [2,1,0], Numbers = [1,2,3,4].

4. starts = [2,1,0], so new_num = min(3*2,2*3,1*5) = 5, so now starts = [2,1,1], Numbers = [1,2,3,4,5].

5. starts = [2,1,1], so new_num = min(3*2,2*3,2*5) = 6, so let us be carefull in this case: we need to increase two numbers from start, because our new number 6 can be divided both by 2 and 3, so now starts = [3,2,1], Numbers = [1,2,3,4,5,6].

6. starts = [3,2,1], so new_num = min(4*2,3*3,2*5) = 8, so now starts = [4,2,1], Numbers = [1,2,3,4,5,6,8]

7. starts = [4,2,1], so new_num = min(5*2,3*3,2*5) = 9, so now starts = [4,3,1], Numbers = [1,2,3,4,5,6,8,9].

8. starts = [4,3,1], so new_num = min(5*2,4*3,2*5) = 10, so we need to update two elements from starts and now starts = [5,3,2], Numbers = [1,2,3,4,5,6,8,9,10]

9. starts = [5,3,2], so new_num = min(6*2,4*3,3*5) = 12, we again need to update two elements from starts, and now starts = [6,4,2], Numbers = [1,2,3,4,5,6,8,9,10,12].

10. starts = [6,4,2], so new_num = min(8*2,5*3,3*5) = 15, we again need to update two elements from starts, and now starts = [6,5,3], Numbers = [1,2,3,4,5,6,8,9,10,12,15].

class Solution {
    public int nthUglyNumber(int n) {
        int[] nums = new int[n];
        nums[0] = 1;
        int index2 = 0, index3 = 0, index5 = 0;
        for (int i = 1; i < n; i++) {
            nums[i] = Math.min(Math.min(nums[index2] * 2, 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];
    }
}