338. Counting Bits

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探讨了如何高效计算从0到给定整数范围内每个数的二进制表示中1的数量，通过动态规划和位操作实现了线性时间复杂度的解决方案。

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问题描述

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.

Credits:
Special thanks to @ syedee for adding this problem and creating all test cases.

题目链接：

思路分析

给一个非零的数，输出从0到这个数的二进制表示中分别有多少个1。

使用i&(i-1)的操作可以使i中的1的个数减1，动态规划的思想，再加上1就可以了，非常巧妙，我自己想不出。

代码

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> res(num + 1, 0);
        for (int i = 1; i <= num; i++){
            res[i] = res[(i-1)&i] + 1;
        }
        return res;
    }
};