338. Counting Bits 二进制中1的个数_i = bits[i] 1;-优快云博客

本文介绍了一种高效计算非负整数范围内每个数二进制表示中1的数量的方法。通过观察数的二进制表示，提出了一种线性时间复杂度的算法，避免了使用内置函数，实现了在单次遍历中完成计算。

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

Example 1:

Input: 2
Output: [0,1,1]

Example 2:

Input: 5
Output: [0,1,1,2,1,2]

Follow up:

It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
Space complexity should be O(n).
Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

题意：求一个数的二进制队列中有多少个1.

思路：

最笨的方法：将数拆分成二进制后，再逐个纪录1的个数。

简单的方法：原数/2等于将二进制队列向右移一位，如4/2=2 ( 100>>1 = 10)。

可见某数的二进制队列为此数除2后的二进制队列再加上0或1。

代码：

class Solution {
    public int[] countBits(int num) {
        int[] count = new int[num+1];
        for(int i=1;i<=num;i++){
            // 'x>>y'表示x的二进制右移y位，等于除以2^y；
            // ‘z&1’表示z与1相与,  0&1=0,1&1=1, 2&1=0
            count[i] = count[i>>1] + (i&1);
        }
        return count;
    }
}