LeetCode(九)231. Power of Two&405. Convert a Number to Hexadecimal

231.Power of Two

题目要求判断一个数是否是2的乘方。
可以使用循环求解，也可以用位运算如下

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

使用如下代码可以求解一个数的二进制中1的个数

 int ans = 0;
    while(x != 0){
        ++ans;
        x = x&(x-1);
    }
    return ans;

405.Convert a Number to Hexadecimal

题目要求如下

Given an integer, write an algorithm to convert it to hexadecimal. For
negative integer, two’s complement method is used.

Note:

All letters in hexadecimal (a-f) must be in lowercase. The hexadecimal
string must not contain extra leading 0s. If the number is zero, it is
represented by a single zero character ‘0’; otherwise, the first
character in the hexadecimal string will not be the zero character.
The given number is guaranteed to fit within the range of a 32-bit
signed integer. You must not use any method provided by the library
which converts/formats the number to hex directly.
Example 1:

Input: 26

Output: “1a”

Example 2:

Input:
-1

Output: “ffffffff”

解法如下

public class Solution {

     char[] map = {'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'};
    public String toHex(int num) {

        if(num==0){
            return "0";
        }
        String result="";
        while(num!=0){
            result=map[(num&15)]+result;
            num=num>>>4;//无符号右移
        }
        return result;
    }
}

处理的过程就是每次获取最低4位，将其对应的16进制表示添加到结果字符串的开头。计算机里面就是二进制补码表示的，我们需要做的，就是把数字表示成32位二进制，然后每四位映射成一个16进制数字，最后去掉前导0。不需要考虑负数补码怎么实现。

延伸Power of Four与上题类似，要去掉2的乘方留下4的乘方
解法如下

public boolean isPowerOfFour(int num) {
        return num > 0 && (num&(num-1)) == 0 && (num & 0x55555555) != 0;
      //0x55555555 十六进制数，是01010101010101010101010101010101
    }