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.

方法1：

public class Leetcode_264_UglyNumberII {

    public static void main(String[] args) {
        Leetcode_264_UglyNumberII leetcode_264_uglyNumberII = new Leetcode_264_UglyNumberII();
        System.out.println(leetcode_264_uglyNumberII.nthUglyNumber(11));
    }

    /**
     * 1 乘以之前確定好的醜數才行，並不是單純的+2，+3，+5，因為14不是醜數
     * 2 如果Min2 == Min3 < Min5 ,則取完最小值後，Min2,Min3都會增加，這樣是正確的，繼續看下一次哪個數最小
     * @param n
     * @return
     */
    //Runtime: 4 ms, faster than 76.19% of Java online submissions for Ugly Number II.
    public int nthUglyNumber(int n) {
        int[] ugly = new int[n];
        int index2 = 0, index3 = 0, index5 = 0, min2 = 2, min3 = 3, min5 = 5;
        ugly[0] = 1;
        for (int i = 1; i < n; i++) {
            ugly[i] = Math.min(Math.min(min2, min3), min5);
            if (ugly[i] == min2) {
                min2 = 2 * ugly[++index2];
            }
            if (ugly[i] == min3) {
                min3 = 3 * ugly[++index3];
            }
            if (ugly[i] == min5) {
                min5 = 5 * ugly[++index5];
            }
        }
        return ugly[n - 1];
    }

    /**
     * 這是錯誤的方法
     * @param n
     * @return
     */
    public int nthUglyNumber2(int n) {
        int[] ugly =new int[n];
        int index2=1,index3=1,index5=1,min2=2,min3=3,min5=5;
        ugly[0]=1;
        for(int i=1;i<n;i++){
            ugly[i] =Math.min( Math.min(min2,min3),min5);
            if( ugly[i] == min2){
                min2 +=2;// 12 +2 = 14 ,14並不是醜數，需要乘以之前確定的醜數才行
            }
            if( ugly[i] == min3){
                min3 +=3;//12 +3 = 15，
            }
            if( ugly[i] == min5){
                min5 +=5;
            }
        }
        return ugly[n-1];
    }
}
 

O(n) Java solution

https://leetcode.com/problems/ugly-number-ii/discuss/69362/O(n)-Java-solution

就是比較*2，*3，*5哪個數是最小的，如果*2的數最小，則繼續使*2的數變大一次，下次再比較誰最小