191. Number of 1 Bits

Link

Solution

 public int hammingWeight(int n) {
        int count = 0;
        while(n != 0){
            n = n & (n-1);
            count++;
        }
        return count;
    }

ticky的方法: 只能列舉來了解是怎麼運作: n = 3 = 11 n-1 = 2 = 10 可以簡化成 n = (3&2) = 1 , count = 1,的問題

n = 4 = 100 n = 3 = 011 可以簡化成 n = (4&3) = 0 , count = 1,的問題