Leetcode 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 theHamming weight).

For example, the 32-bit integer ’11' has binary representation00000000000000000000000000001011, so the function should return 3.

方法一：移位法

          通过与运算符判断最低位/高位是否为1，然后在左移/右移。重复步骤直到原数归为零，无符号数目直接右移即可。

<span style="font-size:14px;">public class Solution {    // you need to treat n as an unsigned value    public int hammingWeight(int n) {        int mask = 0B1;         int count = 0;        while(n != 0B0){            if((n & mask) == 0B1){                count++;            }            n = n>>>1;        }        return count;    }}</span>

方法二：减一相与法，当原数不为0时，将原数与上原数减一再赋给原数。因为每一次减掉1再想与实际上是将左边的1给消去。所以消去几次就有几个1.

<span style="font-size:14px;">public class Solution {    // you need to treat n as an unsigned value    public int hammingWeight(int n) {        int count = 0;        while(n != 0b0){            n = n & (n-1);            count++;        }        return count;    }}</span>