LeetCode--Ugly Number&&Ugly NumberⅡ--JavaScript&Java

Ugly NumberⅡ

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.

JavaScript
/*
* * @param {number} n * @return {number} */ var nthUglyNumber = function(n) { var l1 = new Array(0), l2 = new Array(0), l3 = new Array(0), i = 0, ans = 0; l1[0] = 1; l2[0] = 1; l3[0] = 1; for(i = 0;i < n;i ++){ ans = Math.min(Math.min(l1[0],l2[0]),l3[0]); if(l1[0] === ans){ l1.shift(); } if(l2[0] === ans){ l2.shift(); } if(l3[0] === ans){ l3.shift(); } l1.push(ans * 2); l2.push(ans * 3); l3.push(ans * 5); } return ans; };

Java

public class Solution {  
    public int nthUglyNumber(int n) {  
        int u = 0;  
        List<Integer> l1 = new LinkedList<Integer>();  
        List<Integer> l2 = new LinkedList<Integer>();  
        List<Integer> l3 = new LinkedList<Integer>();  
        l1.add(1);  
        l2.add(1);  
        l3.add(1);  
          
        for(int i=0; i<n; i++) {  
            u = Math.min( Math.min(l1.get(0), l2.get(0)), l3.get(0));  
              
            if(l1.get(0) == u) l1.remove(0);  
            if(l2.get(0) == u) l2.remove(0);  
            if(l3.get(0) == u) l3.remove(0);  
              
            l1.add(u*2);  
            l2.add(u*3);  
            l3.add(u*5);  
        }  
        return u;  
    }  
} 

 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.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

 

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JavaScript
/*
* * @param {number} num * @return {boolean} */ var isUgly = function(num) { var i = 2; var temp = num; if(num <= 0) return false; if(num <= 6) return true; while(i * i <= num){ if(num % i === 0){ if(i > 5) return false; while(num % i === 0) num = num / i; } i ++; } if(num > 5) return false; return true; };

 

posted @ 2015-10-29 23:22  Decmber  阅读(209)  评论(0编辑  收藏  举报