常见的bit操作

Swap variable without temp

using XOR

int x = 10, y = 5;

x = x ^ y; // x become 15
y = x ^ y; // y became 10;
x = x ^ y; // x become 5.

Number of 1 Bits

计算一个无符号数的二进制表达里面有多少位的值是1，也就是计算它的Hamming weight。

做法很直观，就是用这个数不断和一个只有当前位为1的数进行and操作, check whether the result is 0，遍历int的所有32位。

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int count = 0;

        for(int i = 0; i < 32; i++) {
            // 结果不是1
            if((n & (1 << i)) != 0) {
                count++;
            }
        }
        return count;
    }
}