338. Counting Bits**

338. Counting Bits**

​​https://leetcode.com/problems/counting-bits/​​

题目描述

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 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.

C++ 实现 1

做这道题需要了解一个结论: 思路 2: ​​https://leetcode.com/problems/number-of-1-bits/discuss/55106/Python-2-solutions.-One-naive-solution-with-built-in-functions.-One-trick-with-bit-operation​​

2.Using bit operation to cancel a ​​1​​ in each round

Think of a number in binary ​​n = XXXXXX1000​​​, ​​n - 1 is XXXXXX0111​​​. ​​n & (n - 1)​​​ will be ​​XXXXXX0000​​​ which is just cancel the last ​​1​​

也就是说, ​​n & (n - 1)​​​ 的结果是 n 中最右边的 1 变为了 0, 那么持续这样操作下去, 最终 n 就会变为 0. 该思路也可以用来处理 ​​191. Number of 1 Bits*​​ 这道题.

另外也说明, ​​n​​​ 和 ​​n & (n - 1)​​​ 之间的关系是, 前者的二进制表示要比后者在最右边多一个 1. 对于本题, 如果用 ​​f[i]​​​ 表示数 ​​i​​​ 的二进制表示中 ​​1​​ 的个数, 那么有如下递推式:

f[i] = f[i & (i - 1)] + 1

于是代码如下:

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