https://leetcode.com/problems/power-of-three/description/
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?
- 数学题
- 第一种解法，简单直接，循环除以3直到最好余数为1。注意边界值。
- 第二种解法，int最大值在C/C++语言中与机器位长相关，32位下是2^32-1，Java中int是固定的32位，即2^32-1。那么可以算出来int下最大的3的幂为3^19=1162261467。只要能被它整除的必然是3的幂。
- https://leetcode.com/problems/power-of-three/solution/
1 // 2 // main.cpp 3 // LeetCode 4 // 5 // Created by Hao on 2017/3/16. 6 // Copyright © 2017年 Hao. All rights reserved. 7 // 8 9 #include <iostream> 10 using namespace std; 11 12 class Solution 13 { 14 public: 15 bool isPowerOfThree(int n) 16 { 17 if (n < 1) { 18 return false; 19 } 20 21 while (n % 3 == 0) { 22 n /= 3; 23 } 24 25 return n == 1; 26 } 27 28 bool isPowerOfThree2(int n) 29 { 30 return n > 0 && 1162261467 % n == 0; 31 } 32 }; 33 34 int main(int argc, char* argv[]) 35 { 36 Solution testSolution; 37 38 auto nums = {0, 1, 2, 3, 4, 6, 9, 27, 1162261467}; 39 40 for (auto num : nums) 41 { 42 cout << num << " isPowerOfThree ? " << testSolution.isPowerOfThree(num) << endl; 43 cout << num << " isPowerOfThree2 ? " << testSolution.isPowerOfThree2(num) << endl; 44 } 45 46 return 0; 47 }
0 isPowerOfThree ? 0 0 isPowerOfThree2 ? 0 1 isPowerOfThree ? 1 1 isPowerOfThree2 ? 1 2 isPowerOfThree ? 0 2 isPowerOfThree2 ? 0 3 isPowerOfThree ? 1 3 isPowerOfThree2 ? 1 4 isPowerOfThree ? 0 4 isPowerOfThree2 ? 0 6 isPowerOfThree ? 0 6 isPowerOfThree2 ? 0 9 isPowerOfThree ? 1 9 isPowerOfThree2 ? 1 27 isPowerOfThree ? 1 27 isPowerOfThree2 ? 1 1162261467 isPowerOfThree ? 1 1162261467 isPowerOfThree2 ? 1 Program ended with exit code: 0