231. 2的幂

谁践踏了优雅 2023-02-11 04:37 33阅读 0赞

# 题目来源 #

[leetcode][]

# 题目描述 #

相似的题目：[326. 3的幂][326. 3]

[342. 4的幂][342. 4]

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3poaXpoZW5nZ3Vhbg_size_16_color_FFFFFF_t_70]

# 题目解析 #

## 模拟 ##

public static boolean isPowerOfTwo(int n) {
        
            if (n == 1) return true;
            if (n < 1) return false;
            int sum = 0;
            while (n > 1)
            {
        
                sum += n % 2;   //1对2取余，得到1
                n = n / 2;
            }
            return sum == 0;
        }

## 位运算 ##

2 10
    4 100
    8 1000
    16 10000

判断该位的二进制是不是只有1个1

public static boolean isPowerOfTwo(int n) {
        
            return (n > 0) && ((n&(n-1)) == 0);
        }

## 解法3 ##

public static boolean isPowerOfTwo(int n) {
        
            return n > 0 && 0 == 2147483648 % n;
        }

## 错误的解法【为什么】 ##

public boolean isPowerOfTwo(int n) {
        
            double v = Math.log(n) / Math.log(2);
            return (n > 0) && && ((int) v == v); 
        }

public static boolean isPowerOfTwo(int n) {
        
            double v = Math.log(n) / Math.log(2);
            double db = (int)v - v;
            return db >= (-1e-6) && db<= (1e-6);
        }

因为double和float类型在存储时都有误差，判断两个数是否相等，只要两者之差小于一个极小值，即可认为两者相等

[leetcode]: https://leetcode-cn.com/problems/power-of-two/
[326. 3]: https://blog.csdn.net/zhizhengguan/article/details/106272896
[342. 4]: https://blog.csdn.net/zhizhengguan/article/details/106273610
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