<LeetCode OJ> 326. Power of Three

326. Power of Three

Question

Total Accepted: 1159 Total
Submissions: 3275 Difficulty: Easy

推断给定整数是否是3的某次方。

Given an integer, write a function to determine if it is a power of three.

Follow up:

Could you do it without using any loop / recursion?

完毕此题。是否能不用迭代或者循环？(<--这是曾经的错误翻译，羞愧。

正确的应该是：完毕此问题是否能不用不论什么形式的循环和递归？)

分析：DONE

最简单的方法，重复迭代（即循环。可是题目不建议）

简单分析：

a）3^x。无论x正或者负。这个值一定大于0

b）假设一个数是3的x次方那么。重复除以3。终于一定等于1。return true

c）否则不是满足要求的数,return false

时间复杂度：O(lg(n))。空间复杂度：O(1)

class Solution {
public:
    bool isPowerOfThree(int n) {
        int num=n;
        while(num>0 && num%3==0)
            num/=3;
        return num==1;
    }
};

2，和上面的思想一样：递归形式的解法（题目不建议）

时间复杂度：O(lg(n))，空间复杂度：O(lg(n))

class Solution {
public:

    bool isPow3(int n,int step) {
        if(pow(3,step)==n)
                return true;
        if(pow(3,step)>n)
                return false;
        if(pow(3,step)<n)
                return isPow3(n,++step);
    }
    bool isPowerOfThree(int n) {
        int step=0;
        return isPow3(n,step);
    }
};

不用不论什么循环：（參考讨论区）

3。不论什么一个3的x次方一定能被int型里最大的3的x次方整除，例如以下所看到的：

return n>0?

!(1162261467 % n):0;

4，或者直接列举：

由于n是int型整数，所以其内满足要求的数还不到32个（2的x次方才32个）,所以能够直接列举

class Solution {
public:
    bool isPowerOfThree(int n) {
        return (n == 1 || n == 3 || n == 9 || n == 27 || n == 81 || n == 243 || n == 729 || n == 2187 || n == 6561 || n == 19683 || n == 59049 || n == 177147 || n == 531441 || n == 1594323 || n == 4782969 || n == 14348907 || n == 43046721 || n == 129140163 || n == 387420489 || n == 1162261467);
    }
};

5。log函数

一个主要的事实就是假设n是3的x次方，那么以3为低对数后一定是一个整数，否则不是

class Solution {
public:
    bool isPowerOfThree(int n) {
        double res = log10(n) / log10(3);  //有精度问题，不要用以指数2.718为低的log函数
        return (res - int(res) == 0) ?

true : false;
    }
};

附带讨论区一篇原文：

Well, this problem doesn't seem to be quite interesting or worthwhile to think about at a first glance. I had the same feeling at the beginning. However, after seeing a couple of posts, I saw a couple of interesting ways. So here is a summary post and hope
you learn something from others' solutions.

Two trivial solutions first:

Recursive Solution

public boolean isPowerOfThree(int n) {
    return n>0 && (n==1 || (n%3==0 && isPowerOfThree(n/3)));
}

Iterative Solution

update following Stefan's answer below:

public boolean isPowerOfThree(int n) {
    if(n>1)
        while(n%3==0) n /= 3;
    return n==1;
}

my original code: public boolean isPowerOfThree(int n) { while(n>1) { if(n%3!=0) return false; n /= 3; } return n<=0 ? false : true; }

It's all about MATH...

Method 1

Find the maximum integer that is a power of 3 and check if it is a multiple of the given input. (related
post)

public boolean isPowerOfThree(int n) {
    int maxPowerOfThree = (int)Math.pow(3, (int)(Math.log(0x7fffffff) / Math.log(3)));
    return n>0 && maxPowerOfThree%n==0;
}

Or simply hard code it since we know maxPowerOfThree = 1162261467:

public boolean isPowerOfThree(int n) {
    return n > 0 && (1162261467 % n == 0);
}

It is worthwhile to mention that Method 1 works only when the base is prime. For example, we cannot use this algorithm to check if a number is a power of 4 or 6 or any other composite number.

Method 2

If log10(n) / log10(3) returns an int (more precisely, a double but has 0 after
decimal point), then n is a power of 3. (original post). But be
careful here, you cannot use log (natural log) here, because it will generate
round off error for n=243. This is more like a coincidence. I mean when n=243,
we have the following results:

log(243) = 5.493061443340548    log(3) = 1.0986122886681098
   ==> log(243)/log(3) = 4.999999999999999

log10(243) = 2.385606273598312    log10(3) = 0.47712125471966244
   ==> log10(243)/log10(3) = 5.0

This happens because log(3) is actually slightly larger than its true value due
to round off, which makes the ratio smaller.

public boolean isPowerOfThree(int n) {
    return (Math.log10(n) / Math.log10(3)) % 1 == 0;
}

Method 3 related
post

public boolean isPowerOfThree(int n) {
    return n==0 ? false : n==Math.pow(3, Math.round(Math.log(n) / Math.log(3)));
}

Method 4 related post

public boolean isPowerOfThree(int n) {
    return n>0 && Math.abs(Math.log10(n)/Math.log10(3)-Math.ceil(Math.log10(n)/Math.log10(3))) < Double.MIN_VALUE;
}

Cheating Method

This is not really a good idea in general. But for such kind of power questions,
if we need to check many times, it might be a good idea to store the desired powers into an array first. (related
post)

public boolean isPowerOfThree(int n) {
    int[] allPowerOfThree = new int[]{1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467};
    return Arrays.binarySearch(allPowerOfThree, n) >= 0;
}

or even better with HashSet:

public boolean isPowerOfThree(int n) {
    HashSet<Integer> set = new HashSet<>(Arrays.asList(1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969, 14348907, 43046721, 129140163, 387420489, 1162261467));
    return set.contains(n);
}

New Method Included at 15:30pm Jan-8th

Radix-3 original post

The idea is to convert the original number into radix-3 format and check if it is of format 10*where 0* means k zeros
with k>=0.

public boolean isPowerOfThree(int n) {
    return Integer.toString(n, 3).matches("10*");
}

注：本博文为EbowTang原创，兴许可能继续更新本文。假设转载。请务必复制本条信息。

原文地址：http://blog.csdn.net/ebowtang/article/details/50485622

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895