LeetCode_Bit Manipulation

231. Power of Two

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

   

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

191. Number of 1 Bits

    Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight).

   For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should return 3.

//我的解法
class Solution {
public:
    int hammingWeight(uint32_t n) {
        int count = 0;
        while(n != 0)
        {
            if(n%2)
            {count++;}
            n = n>>1;
        }
        return count;
    }
};

//网上大神的解法
int CountOne(unsigned long n)
{
    //0xAAAAAAAA，0x55555555分别是以“1位”为单位提取奇偶位
    n = ((n & 0xAAAAAAAA) >> 1) + (n & 0x55555555);
    //0xCCCCCCCC，0x33333333分别是以“2位”为单位提取奇偶位
    n = ((n & 0xCCCCCCCC) >> 2) + (n & 0x33333333);
    //0xF0F0F0F0，0x0F0F0F0F分别是以“4位”为单位提取奇偶位
    n = ((n & 0xF0F0F0F0) >> 4) + (n & 0x0F0F0F0F);
    //0xFF00FF00，0x00FF00FF分别是以“8位”为单位提取奇偶位
    n = ((n & 0xFF00FF00) >> 8) + (n & 0x00FF00FF);
    //0xFFFF0000，0x0000FFFF分别是以“16位”为单位提取奇偶位
    n = ((n & 0xFFFF0000) >> 16) + (n & 0x0000FFFF);

    return n;
}

参考： http://blog.csdn.net/yunyu5120/article/details/6692072

190. Reverse Bits

   Reverse bits of a given 32 bits unsigned integer.

   For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary        as  00111001011110000010100101000000).

  

//我的解法
class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t temp[32];
        uint32_t n1 = 0;
        for(int i = 0;i<32;i++)
        {
            temp[i] = n % 2;
            n = n >> 1;
        }
        for(int i = 0;i<32;i++)
        {
            n1 += (temp[i]<<(31-i));
        }
        return n1;
    }
};

//网上大神的写法
class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t value = 0;
        uint32_t mask = 1;
        for (uint32_t i = 0; i < 32; ++i) {
            value = (value<<1 )|((n&mask)>>i);
            mask <<=1;
        }
        return value;
    }
};

参考：http://blog.csdn.net/feliciafay/article/details/44536827

136. Single Number

Given an array of integers, every element appears twice except for one. Find that single one.

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int result = 0;
        for(int i = 0;i < nums.size();i++)
        {
            result = result ^ nums[i];
        }
        return result;
    }
};

137. Single Number two

Given an array of integers, every element appears three times except for one. Find that single one.

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        vector<int> flag(32);
        int result = 0;
        for(int k = 0;k < 32;k++)
        {
            for(int i = 0;i < nums.size();i++)
            {
                flag[k] += ((nums[i]>>k) & 1);
            }
        }
        for(int k = 0;k < 32;k++)
        {
            result += ((flag[k] % 3)<<k);
        }
        return result;
    }
};

260. Single Number three

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

class Solution {
public:
    vector<int> singleNumber(vector<int>& nums) {
        vector<int> result(2);
        int temp = 0;
        for(int i = 0;i < nums.size();i++)
        {
            temp ^= nums[i];            
        }
        int flag = 0;
        while(!(temp & 1))
        {
            temp >>= 1;
            flag++;
        }
        for(int i = 0;i < nums.size();i++)
        {
            int tmp = nums[i]>>flag;
            if(tmp & 1)
                result[0] ^= nums[i];
            else
                result[1] ^= nums[i];
        }
        return result;
    }
};