統計一個Byte中1的個數，演算法儘可能高效能

/*統計一個Byte中1的個數，演算法儘可能高效能*/#include <iostream>using namespace std;//方法1：對2模數得出最後一位，然後除以2實現右移1位，迴圈8次int count1(unsigned char c){int cnt=0;for (int i=0;i!=8;++i){cnt+=c%2;c=c/2;}return cnt;}//方法2：對0x01執行&操作，統計最右邊一位的1，然後右移1位，迴圈8次，o(n)複雜度int count2(unsigned char c){int cnt=0;for (int i=0;i!=8;++i){cnt+=(c&0x01);c>>=1;}return cnt;}//方法3：c=c&(c-1),只統計c中1的個數，迴圈次數是位元組中1的個數，效率比方法2高int count3(unsigned char c){int cnt=0;while (c!=0){c&=(c-1);++cnt;}return cnt;}//方法4：查表法，建立一個含有256個整數的數組，每個整數值代表的是c所在的位置含有1的個數//時間複雜度o(1),空間複雜度o(2^n)int countTable[256]={ 0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8};int count4(unsigned char c){return countTable[c];}//測試程式int main(){char ch='a';cout<<count1(ch)<<endl;cout<<count2(ch)<<endl;cout<<count3(ch)<<endl;cout<<count4(ch)<<endl;return 0;}