231,338,326,Power of Two(Three,Four)Counting Bits 类似的技巧题

231. Power of Two

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

题意：

一个数是否 2的n次方的判断

思想：

 把该数n转换为2进制，与该数n-1进行&操作 可以判断出是否为2^n

C++ AC代码：Time O(1) Space O(1)

class Solution {public:    bool isPowerOfTwo(int n) {       if(n<=0) return false;        return !(n&(n-1));    }};

338. Counting Bits

该题为231的一个引申题

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

Example:
For num = 5 you should return [0,1,1,2,1,2].

Follow up:

• It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
• Space complexity should be O(n).
• Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

题意：

给出一个数num 求0-num中每个数转化为2进制数 1的个数。

思想：

可以这样想，假设一个数A 它会是数 B=A&A-1的2进制1的个数加1 如3 2进制(11) 会是2 2进制(10)中2进制个数加1 

C++ AC代码： Time O(n) Space(n)

class Solution {public:    vector<int> countBits(int num) {        vector<int> res(num+1,0);        for(int i=1;i<=num;i++){            res[i] =res[i&(i-1)] + 1;        }        return res;    }};

326. Power of Three

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

题意：

一个数是否 3的n次方的判断

思想：

方法一：

假设这个数最大为int型范围内2^31-1 那么先求3的^n能到达这个int型范围内的最大值 3^19 max

直接判断n>0&&max%n==0

方法二：

fmod(log10(num)/log10(3),1)==0;fmod c++判断模运算

C++ AC代码：Time O(1) Space O(1)

public class Solution {    public boolean isPowerOfThree(int n) {        return n > 0 && 1162261467 % n == 0; //3^19 = 1162261467    }}

342. Power of Four

Given an integer (signed 32 bits), write a function to check whether it is a power of 4.

题意：

一个数是否 4的n次方的判断

思想：  

方法一：

fmod(log10(num)/log10(4),1)==0

方法二：先判断是否是2^n 在判断该数a-1是否整除3

return num>0&&(num&(num-1))==0&&((num-1)%3==0);

C++ AC代码： Time O(1)  Space(1)

class Solution {public:    bool isPowerOfFour(int num) {        return num>0&&(num&(num-1))==0&&((num-1)%3==0);    }};