leetcode 338. Counting Bits-优快云博客

本文介绍了一种使用位运算高效计算一个整数范围内所有数的二进制表示中1的数量的方法。通过巧妙地利用位操作特性，该算法能够在线性时间内完成计算，避免了传统方法中的高时间复杂度问题。

Total Accepted: 76419
Total Submissions: 125855
Difficulty: Medium
Contributor: LeetCode
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:
For num = 5 you should return [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.

关键是位运算
自己的low鸡写法

public class Solution {
    public int[] countBits(int num) {
        int[] res = new int[num+1];
        res[0] = 0;
        if(num==0) return res;
        res[1] = 1;
        if(num==1) return res;
        for(int i=1; i<=32;i++){
            // int offset = (int)Math.pow(2,i);
            int offset = 1 << i;
            for(int j= offset; j< 1 << i+1; j++){
                res[j] = res[j-offset]+1;
                if(j==num) return res;
            }
        }
        return res;
    }
}

发现可以通过移位来做，公式为：
$f[i] = f[i / 2] + i \% 2.$

public int[] countBits(int num) {
    int[] f = new int[num + 1];
    for (int i=1; i<=num; i++) f[i] = f[i >> 1] + (i & 1);
    return f;
}