[Leetcode] Power of Three

最新推荐文章于 2019-03-02 10:35:21 发布

Jasonfang0118

最新推荐文章于 2019-03-02 10:35:21 发布

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本文介绍两种方法来判断一个整数是否为3的幂。第一种方法使用除数法，通过不断除以3直到无法整除，检查最终结果是否为1。第二种方法利用整型数特点，通过计算最大可能的3的幂值（1162261467），检查该值是否能被待判断的数整除。

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

判断一个数是否为3的幂：

一，除数法，最容易想到的方法：

1. 首先是能够整除

2. 整除到最后的数一定为1则说明TRUE否则FALSE

bool isPowerOfThree(int n) {
    if (n <= 0)
        return false;
    
    while(0 == (n % 3)){
        n/=3;
    }    
    
    return (n == 1);
}

二，根据整型数的特点，总结规律

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

引用从解答中的分析：

MaxInt = $\frac{ 2^{32} }{2} - 1$ since we use 32 bits to represent the number, half of the range is used for negative numbers and 0 is part of the positive numbers

Knowing the limitation of n, we can now deduce that the maximum value of n that is also a power of three is 1162261467. We calculate this as:

3^{\lfloor{}log_3{MaxInt}\rfloor{}} = 3^{\lfloor{}19.56\rfloor{}} = 3^{19} = 1162261467

Therefore, the possible values of n where we should return true are $3^0$ , $3^1$ ... $3^{19}$ . Since 3 is a prime number, the only divisors of $3^{19}$ are $3^0$ , $3^1$ ... $3^{19}$ , therefore all we need to do is divide $3^{19}$ by n. A remainder of 0 means n is a divisor of $3^{19}$ and therefore a power of three.