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[leetcode]ugly number i ii

admin 11月 11, 2020 0

Ugly Number I

题目描述

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.

Note that 1 is typically treated as an ugly number.

代码

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public class Solution {
    
    public boolean (int num) {
        if(num <= 0) return false;
        if(num == 1) return true;
        while(num % 2 == 0) num = num / 2;
        while(num % 3 == 0) num = num / 3;
        while(num % 5 == 0) num = num / 5;
        return num == 1;
    }
}

Ugly Number II

题目描述

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.

Hint:

1.The naive approach is to call isUgly for every number until you reach the nth one. Most numbers are not ugly. Try to focus your effort on generating only the ugly ones.
2.An ugly number must be multiplied by either 2, 3, or 5 from a smaller ugly number.
3.The key is how to maintain the order of the ugly numbers. Try a similar approach of merging from three sorted lists: L1, L2, and L3.
4.Assume you have Uk, the kth ugly number. Then Uk+1 must be Min(L1 * 2, L2 * 3, L3 * 5).

代码

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public class Solution {
    
    //ugly[1] = min(2*ugly[0], 3*ugly[0], 5*ugly[0])
    //ugly[2] = min(2*ugly[1], 3*ugly[0], 5*ugly[0])
    //
    public int nthUglyNumber(int n) {
        if(n <= 0) return 0;
        
        int[] ugly = new int[n];
        ugly[0] = 1;
        int index2 = 0, index3 = 0, index5 = 0; 
        for(int i = 1; i < n;i++){
            ugly[i] = Math.min(Math.min(2 * ugly[index2], 3 * ugly[index3]), 5 * ugly[index5]);
            if(ugly[i] == 2 * ugly[index2]) index2++;
            if(ugly[i] == 3 * ugly[index3]) index3++;
            if(ugly[i] == 5 * ugly[index5]) index5++;
        }
        
        return ugly[n-1];
    }
}

Super Ugly Number

题目描述

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.

Note:
(1) 1 is a super ugly number for any given primes.
(2) The given numbers in primes are in ascending order.
(3) 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.

代码

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public class Solution {
    //思路类似，Ugly Number II，只是质数因子不是固定的三个，由primes数组提供，
    //则我们维护一个长度跟primes数组长度一样indexes数组即可
    public int nthSuperUglyNumber(int n, int[] primes) {
        if(n <= 0) return 0;
        
        int[] ugly = new int[n];
        ugly[0] = 1;
        
        int[] indexes = new int[primes.length];
        
        for(int i = 1; i < n; i++){
            ugly[i] = Integer.MAX_VALUE;
            for(int j = 0; j < primes.length; j++){
                ugly[i] = Math.min(ugly[i], primes[j]*ugly[indexes[j]]);
            }
            
            for(int j = 0; j < primes.length; j++){
                if(ugly[i] == primes[j]*ugly[indexes[j]]){
                    indexes[j]++;
                }
            }
        }
        
        return ugly[n-1];
    }
}
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