LeetCode 338. 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.

解析：根据给定的整数，对于 0 ≤ i ≤ num 范围中的每个数字 i ，计算其二进制数中的 1 的数目并将它们作为数组返回。

思路：首先想到的是遍历每一个数值，然后计算当前数字中包含1的个数。这里要求空间复杂度和时间复杂度均为O(N)，争取一次遍历即可获取结果。因此考虑动态规划的思路，计算当前数字结果时尽量用到之前的结果，分析二进制数值的特点，发现如果当前数字i是偶数，则i是由i/2左移得到的，所以i包含1的个数与i/2中1的个数相同；如果i是奇数，则在i-1基础上增加1.具体代码如下：

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