Ugly Number的三道题

263. Ugly Number

问题：

ugly number 是只可以被2， 3， 5整除的数，特殊的是，1也是一个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 

问题：

找到第n个ugly number

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

假设p1,p2,p3分别指向最后一个被2乘过，最后一个被3乘过，最后一个被5乘过的下标。

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

因为dp中的每一个数字，都是要被2，3，5乘上一遍的。这三个指针开始都指向dp[0]，谁被用过一次，就把谁往前移动一格。

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

问题：

和上一道题一样，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：

和上面一道题一样，有多少个primes就keep多少个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];    }}