264.Ugly Number II(Medium)

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 n^{th 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: L_{1, L2, and L3.}
4. Assume you have U_{k, the k^{th ugly number. Then Uk+1 must be Min(L1 * 2, L2 * 3, L3 * 5).}}

　　题意就不翻译了，意思就是把丑数从小到大排列，输出制定位置的那个丑数是多少，按照上一题的那种朴素判断当然是行得通的，只要把1~n的数全部都验证一遍，自然可以找到第n个，但是认真想想，一旦n大了之后，其实丑数的个数是不多的，也就是说验证成功的次数远远小于失败的，所以用朴素的验证方式，显然是相当浪费时间（事实上也是），所以我们的思路就是想到如何一个一个计算出丑数，找到其中的规律，主动去计算丑数而不是被动得验证是不是丑数

My Solution

(Java) Version 1 Time: 131ms:

　　计算丑数的原理当然不难，不要想到这一点还是花了不少功夫，首先每个丑数都是由2,3,5相乘得来的，而且每一个相同乘数的个数还不定，比如有5个5，3个1之类的，所以只能从小到大一个一个算出来，然后我们又发现了每个丑数其实都是由前面小的丑数乘以2,3,5得来的，那就很明显了，我们只要把前面的丑数都乘以2,3,5，然后结果里面最小的那个就一定是下一个丑数了

public class Solution {
    public int nthUglyNumber(int n) {
        int[] nums = new int[n];
        nums[0]=1;
        int max2=0,max3=0,max5=0,index;
        for(int i=0;i

(Java) Version 2 Time: 46ms (By luke.wang.7509):

　　这个解法的生成形式和上一个并无不同，不过上一个解法每一次都要从第一个丑数开始乘显然是没有必要的，因为有很多重复，这里就避免了这个问题

public class Solution {
    public int nthUglyNumber(int n) {
        if(n<=0) return 0;
        int a=0,b=0,c=0;
        List table = new ArrayList();
        table.add(1);
        while(table.size()

(Java) Version 3 Time: 4ms (By TheKingRingReal):

　　DFS是深度优先的搜索算法，这里我还没能理解，不过从结果上看，已经很显然了……4ms快到妈都不认得是谁
　　作者的解释：

we can get a tree like this which contains all the ugly number ;just use three point to dfs because of I can not find the relationship between three point ; so we can think it just like this picture

each point 3 and 5 can have the 2_3and 2_5;3_5 child.and we can use a tricky in programming that if we meet 2_3=6 we should p2++;p3++;if 35:p5++;p3++;just like the code writ in DFS function*

public class Solution {
    public int nthUglyNumber(int n) {
            int[] dp=new int[n];dp[0]=1;
            return DFS(dp,1,0,0,0,n);
    }

    private int DFS(int[] dp, int i, int p2, int p3, int p5, int n) {
        if (i==n)return dp[n-1];
        dp[i]=Math.min(dp[p2]*2, Math.min(dp[p3]*3,dp[p5]*5));
        if (dp[i]==dp[p2]*2)p2++;
        if(dp[i]==dp[p3]*3)p3++;
        if (dp[i]==dp[p5]*5)p5++;
        return DFS(dp, i+1, p2, p3, p5, n);
    }
}

(Java) Version 4 Time: 2ms (By zmajia):

　　这个哈哈哈哈哈哈逗逼把测试样例的极限试了出来，然后打表了哈哈哈哈哈哈，实在是太坏了，所以获得了最快的速度，已知极限用打表的方式空间换时间还是很管用的

public class Solution {
        static int[] save = new int[3000];
        static int flag = 0;
        public int nthUglyNumber(int n) {
           if(flag == 0){
               flag ++;
               init();
           }
           return save[n-1];
        }
        public static void init(){
            int _2 = 0, _3 = 0, _5 = 0;
            save[0] = 1;
            for(int i = 1; i < 3000; i++){
                int min = Math.min(save[_2]*2, Math.min(save[_3]*3,save[_5]*5));
                if(save[_2]*2 == min) {
                    save[i] = save[_2]*2;
                    _2 ++;
                }
                if(save[_3]*3 == min) {
                    save[i] = save[_3]*3;
                    _3 ++;
                }
                if(save[_5]*5 == min) {
                    save[i] = save[_5]*5;
                    _5 ++;
                }
            }
        }
    }

(Java) Version 5 Time: 108ms (By saharH):

　　TreeSet我没有怎么研究过，虽然慢但这是我看到的最简洁的做法，其中起作用的应该就是TreeSet的一些特性了，值得研究

public class Solution {
    public int nthUglyNumber(int n) {
        TreeSet hs = new TreeSet<>(Arrays.asList(1l, 2l, 3l, 5l));
        Long e = 1l;
        while( n != 0){
            e = hs.pollFirst();
            hs.add(e * 2);
            hs.add(e * 3);
            hs.add(e * 5);
            n--;
        }
        return e.intValue();
    }
}