p338_counting_bits

p338_counting_bits

1. Method 1: Dynamic Programming

   i run from 0 to num. So when calculating results[i], we have known results[i >> 1]. The difficult of number of 1's is only decided by last bit of results[i].

   results[i] = results[i >> 1] + (i as i32 & 1);

2. Method 2: call function i32.count_ones()

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1. 方法1：动态规划

   i从0运行到num。所以计算results[i]时，我们已经知道results[i >> 1]。1's的个数差只由results[i]最后一个比特决定。

   results[i] = results[i >> 1] + (i as i32 & 1);

2. 方法2：调用函数i32.count_ones()