[LeetCode] (medium) 338. Counting Bits

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本文介绍了一种高效算法，用于计算从0到指定整数范围内每个数字的二进制表示中1的个数，并通过实例展示了算法的具体实现。算法采用线性时间复杂度O(n)，避免使用内置函数，挑战自我提升。

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

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.

找规律

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> result(num+1);
        result[0] = 0;
        if(num == 0) return result;
        result[1] = 1;
        int idx = 0;
        int thre = 2;
        for(int i = 2; i <= num; ++i, ++idx){
            if(i == thre){
                idx = 0;
                thre <<= 1;
            }
            result[i] = result[idx]+1;
            // cout << result[i] << endl;
        }
        return result;
        
    }
};