LeetCode.263(264) UglyNumber&UglyNumber II

题目263:

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.

分析：

class Solution {    public boolean isUgly(int num) {        //判断一个数是否是丑数（只能被2、3、5的正数），注意：1属于典型的丑数.        if(num<=0)return false;        while(num!=1){            if(num%5==0){                num/=5;            }else if(num%3==0){                num/=3;            }else if(num%2==0){                num/=2;            }else{                return false;            }        }        return true;    }}

题目264：

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, and n does not exceed 1690.

分析：

class Solution {    public int nthUglyNumber(int n) {        //找出第n个丑数(只能被2、3、5整除的正数)，1是典型的丑数        //注意：不能使用逐个求出丑数，容易超时（n不超过1690即为Integer.MAX_VALUE）        int [] uglyNumber=new int[n];        //规律为：the first one, which is 1. Then        //k[1] = min( k[0]x2, k[0]x3, k[0]x5). The answer is k[0]x2. So we move 2's pointer to 1. Then we test:        //k[2] = min( k[1]x2, k[0]x3, k[0]x5). And so on.        int index2=0,index3=0,index5=0;        uglyNumber[0]=1;        for(int i=1;i<n;i++){            uglyNumber[i]=Math.min(uglyNumber[index2]*2,Math.min(uglyNumber[index3]*3,uglyNumber[index5]*5));                        //找出最小的uglyNumber            if(uglyNumber[i]==uglyNumber[index2]*2)index2++;            if(uglyNumber[i]==uglyNumber[index3]*3)index3++;            if(uglyNumber[i]==uglyNumber[index5]*5)index5++;        }        return uglyNumber[n-1];    }}