三種比較好的方法:
(1)複雜度為
int test(int n)
{
int count=0;
while(n != 0){
count += n&1;
n >>= 1;
}
return count;
}
(2)
int test(int n)
{
int count=0;
while(n != 0){
n &= n-1;
count ++;
}
return count;
}
(3)
int test(int n)
{
n = (n&0x55555555) + ((n>>1)&0x55555555);
n = (n&0x33333333) + ((n>>2)&0x33333333);
n = (n&0x0f0f0f0f) + ((n>>4)&0x0f0f0f0f);
n = (n&0x00ff00ff) + ((n>>8)&0x00ff00ff);
n = (n&0x0000ffff) + ((n>>16)&0x0000ffff);
return n;
}
比如這個例子,143的二進制表示是10001111,這裡隻有8位,高位的0怎麼進行與的位運算也是0,是以隻考慮低位的運算,按照這個算法走一次
+---+---+---+---+---+---+---+---+
| 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | <---143
+---+---+---+---+---+---+---+---+
| 0 1 | 0 0 | 1 0 | 1 0 | <---第一次運算後
+-------+-------+-------+-------+
| 0 0 0 1 | 0 1 0 0 | <---第二次運算後
+---------------+---------------+
| 0 0 0 0 0 1 0 1 | <---第三次運算後,得數為5
+-------------------------------+
基本思路:先算出相鄰兩位1的個數,然後在算出相鄰四位的個數,遞推到最後32位。