Ugly Number的三道題

263. Ugly Number

問題：

ugly number 是只可以被2， 3， 5整除的數，特殊的是，1也是一個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.

Note that 1 is typically treated as an ugly number.

程序：

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

264. Ugly Number II 

問題：

找到第n個ugly number

Write a program to find the n-th ugly number.
思路：

假設p1,p2,p3分別指向最後一個被2乘過，最後一個被3乘過，最後一個被5乘過的下標。

dp[n] = min(dp[p1]*2, dp[p2]*3, dp[p3]*5)

因爲dp中的每一個數字，都是要被2，3，5乘上一遍的。這三個指針開始都指向dp[0]，誰被用過一次，就把誰往前移動一格。

程序：

public class Solution {
    public int nthUglyNumber(int n) {
    	int[] dp = new int[n];
    	dp[0] = 1;
    	int p1 = 0, p2 = 0, p3 = 0;
    	for(int i = 1; i < n; i++) {
    		dp[i] = Math.min(dp[p1]*2, Math.min(dp[p2]*3, dp[p3]*5));
    		if(dp[i] == dp[p1]*2) p1++;
    		if(dp[i] == dp[p2]*3) p2++;
    		if(dp[i] == dp[p3]*5) p3++;
    		
    	}
    	return dp[n-1];
    }
}

313. Super Ugly Number

問題：

和上一道題一樣，2,3,5改成了primes

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.

思路：

和上面一道題一樣，有多少個primes就keep多少個pointer

程序：

public class Solution {
    public int nthSuperUglyNumber(int n, int[] primes) {
    	int[] index = new int[primes.length];
    	int[] dp = new int[n];
    	dp[0] = 1;
    	for(int i = 1; i < n; i++) {
    		int minvalue = Integer.MAX_VALUE;
    		int pos = -1;
    		for(int k = 0; k < primes.length; k++) {
    			if(dp[index[k]]*primes[k] < minvalue) {
    				minvalue = dp[index[k]]*primes[k];
    				pos = k;
    			}
    		}
    		index[pos]++;
    		dp[i] = minvalue;
    	}
    	return dp[n-1];
    }
}