LeetCode 264. Ugly Number II

原题俩接在这里：https://leetcode.com/problems/ugly-number-ii/

题目：

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.

题解：

找出第n个ugly number.

(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, …

Let arr[i] denotes ith ugly number, it is choosen from previous ugly number * 2, * 3,  * 5.

Thus we need to maintain the index of previous * 2, * 3, * 5 positions. And move corresponding position when it is chosen. 

Note: when 2 numbers are the same, like arr[i2] = 3, arr[i3] = 2. arr[i2] * 2 == arr[i3] * 3. We need to move both i2 and i3. Since we don't want another 6.

Time Complexity: O(n). Space: O(n).

AC Java:

class Solution {
    public int nthUglyNumber(int n) {
        int [] arr = new int[n];
        arr[0] = 1;
        int i2 = 0;
        int i3 = 0;
        int i5 = 0;
        
        int f2 = 2;
        int f3 = 3;
        int f5 = 5;
        
        for(int i = 1; i < n; i++){
            arr[i] = Math.min(f2, Math.min(f3, f5));
            if(arr[i] == f2){
                i2++;
                f2 = arr[i2] * 2;
            }
            
            if(arr[i] == f3){
                i3++;
                f3 = arr[i3] * 3;
            }
            
            if(arr[i] == f5){
                i5++;
                f5 = arr[i5] * 5;
            }
        }
        
        return arr[n - 1];
    }
}

类似Ugly Number.

跟上Super Ugly Number.