191. Number of 1 Bits

Posted 2020-08-08 Machelsky

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

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

Solution 1:

看起来是一道非常简单的题目，但是自己在做的时候却漏洞百出。

思路：存一个count，每次将n向右移一位，检查最后一位是否为1，是1就将count加一。

注意shift number不会改变number本身，要赋值给number本身才可以。

注意overflow的情况，所以不能把n放在while的条件里面做判断。

Bit Manipulation Ref:

http://www.tutorialspoint.com/java/java_basic_operators.htm

 

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {

        int count=31;
        int res=0;
        while(count>=0)
        {
            if((n&1)!=0)
            {
                res++;
            }
            n=n>>>1;
            count--;
        }
        return res;
    }
}

Solution2:

看了discussion，居然有种作弊的方法！！！

return Integer.bitCount(n);

Solution3: 

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int count = 0;
        int mask = 1;
        int i = 0;
        while(i < 32) {
            if ((n&mask) != 0) {
                count++;
            }
            mask <<= 1;
            i++;
        }
        return count;
    }
}

思路跟1是一样的，但是这种是每次移i次，。移位的简写方式：<<=,>>=. >>>三个箭头是fill left by 0.

Solution 4: 

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int count = 0;
        
        for (count = 0; n != 0; count++) {
            n &= n - 1;     // clear the least significant bit set
        }
        
        return count;
    }
}

这解法太巧了！把最右边的1清零，有多少1就清零多少次。然后累加次数

111000 & 110111＝110000

110000 & 101111 ＝100000

100000 & 011111 ＝000000