Ugly Number | & || & |||

Posted 2020-08-03 北叶青藤

Ugly Number

Write a program to check whether a given number is an ugly number`.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 6, 8 are ugly while 14 is not ugly since it includes another prime factor 7.

 Notice

Note that 1 is typically treated as an ugly number.

Example

Given num = 8 return true
Given num = 14 return false

分析：

根据ugly number的定义，不难想出答案。

 1 public class Solution {
 2     /**
 3      * @param num an integer
 4      * @return true if num is an ugly number or false
 5      */
 6     public boolean isUgly(int num) {
 7         if (num <= 0) return false;
 8         
 9         while(num % 5 == 0) {
10             num = num / 5;
11         }
12         
13         while(num % 3 == 0) {
14             num = num / 3;
15         }
16         
17         while(num % 2 == 0) {
18             num = num / 2;
19         }
20         
21         return num == 1;
22     }
23 }

Ugly Number II

Ugly number is a number that only have factors 2, 3 and 5.

Design an algorithm to find the nth ugly number. The first 10 ugly numbers are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12...

 Notice

Note that 1 is typically treated as an ugly number.

Example

If n=9, return 10.

分析：

 1 class Solution {
 2     /**
 3      * @param k: The number k.
 4      * @return: The kth prime number as description.
 5      */
 6     public int nthUglyNumber(int k) {
 7         if (k < 1)
 8             return 0;
 9         int i2 = 0, i3 = 0, i5 = 0;
10         int[] arr = new int[k];
11         arr[0] = 1;
12 
13         int index = 1;
14         while (index < arr.length) {
15             int v2 = arr[i2] * 2;
16             int v3 = arr[i3] * 3;
17             int v5 = arr[i5] * 5;
18 
19             int min = min(v2, v3, v5);
20             arr[index] = min;
21             
22             // cannot use if-else if-else structure because in some cases, v2, v3 v5 can be the same.
23             if (min == v2) {
24                 i2++;
25             }
26             if (min == v3) {
27                 i3++;
28             }
29             if (min == v5) {
30                 i5++;
31             }
32             index++;
33         }
34         return arr[arr.length - 1];
35     }
36 
37     int min(int m1, int m2, int m3) {
38         return Math.min(m1, Math.min(m2, m3));
39     }
40     
41 
42 };

Super Ugly Number

Write a program to find the nth super ugly number.

Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes of size k. For example,[1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32] is the sequence of the first 12 super ugly numbers given primes = [2, 7, 13, 19] of size 4.

 Notice

• 1 is a super ugly number for any given primes.
• The given numbers in primes are in ascending order.
• 0 < k ≤ 100, 0 < n ≤ 10^6, 0 < primes[i] < 1000

Example

Given n = 6, primes = [2, 7, 13, 19] return 13

分析：

原理同上，这次只是改成了数组，而不是3个数。

 1 public class Solution {
 2     /**
 3      * @param n a positive integer
 4      * @param primes the given prime list
 5      * @return the nth super ugly number
 6      */
 7 
 8     public int nthSuperUglyNumber(int k, int[] primes) {
 9         if (k < 1)
10             return 0;
11         int[] indexes = new int[primes.length];
12         int[] values = new int[primes.length];
13         int[] arr = new int[k];
14         arr[0] = 1;
15 
16         int index = 1;
17         while (index < arr.length) {
18             for (int i = 0; i < indexes.length; i++) {
19                 values[i] = primes[i] * arr[indexes[i]];
20             }
21 
22             int min = getMin(values);
23             arr[index] = min;
24 
25             for (int i = 0; i < values.length; i++) {
26                 if (values[i] == min) {
27                     indexes[i]++;
28                 }
29             }
30             index++;
31         }
32         return arr[arr.length - 1];
33     }
34 
35     public int getMin(int[] arr) {
36         int min = arr[0];
37         for (int k : arr) {
38             min = Math.min(min, k);
39         }
40         return min;
41     }
42 }