【编程之美】求二进制数中1的个数

1 题目

对一字节的无符号变量，求其二进制表示中1的个数

2 代码

#include <cstdio>
#include <iostream>
using namespace std;
typedef unsigned char byte;
//一般解法
byte solution1(byte n){
    int cnt = 0;
    while(n){
        if(n%2)
            cnt++;
        n /= 2;
    }
    return cnt;
}
/*
 * 1.最后一位是1，则n&(n-1)最后一个1变为0。
 * 2.最后一位是0，则n&(n-1)将最后一个1变为0。
 * 总之每运算一次n&(n-1)就把最后一个1变为0.
 */
byte solution2(byte n){
    int cnt = 0;
    while(n){
        n &= (n-1);
        cnt ++;
    }
    return cnt;
}

//以空间换时间
byte solution3(int n){
    byte 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    
        };
    return countTable[n];
}
int main(int argc, char *argv[])
{
    byte n = 123;
    printf("%d\n", solution1(n));
    printf("%d\n", solution2(n));
    printf("%d\n", solution3(n));
    return 0;
}

Author: visaya fan <visayafan[AT]gmail.com>

Date: 2011-08-24 00:38:47

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