326. Power of Three
判断给定整数是否是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?
完成此题，能否不用迭代或者循环？(<--这是以前的错误翻译，惭愧。
正确的应该是：完成此问题能否不用任何形式的循环和递归？)
分析：DONE
最简单的方法，反复迭代（即循环，但是题目不建议）
简单分析：
a）3^x，不管x正或者负，这个值一定大于0
b）如果一个数是3的x次方那么，反复除以3，最终一定等于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)
不用任何循环：（参考讨论区）
3，任何一个3的x次方一定能被int型里最大的3的x次方整除，如下所示：
return n>0?!(1162261467 % n):0;
4，或者直接列举：
因为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）,所以可以直接列举
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
一个基本的事实就是如果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 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，后续可能继续更新本文。如果转载，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