LeetCode 338. Counting Bits

【题目】

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.

【题解】

我们先拿num=3为例，结果为[0, 1, 1, 2]，得到3以内数字的结果之后，4~7的结果可以在此基础上确定，如下0~3二进制表示为：

   bin  num

000    0

001    1

010    1

011    2

4~7二进制表示为：

  bin num

100    1

101    2

110    2

111    3

从上面分析可以看出，4~7除了最高位其他位和0~3是相同的，而最高位是1，这样4~7的结果就可以在0~3的基础上分别加1获得，题目就很容易解决了。

【代码】

    vector<int> countBits(int num) {        vector<int> result(num + 1, 0);        if (num == 0)            return result;                    result[1] = 1;        int nCount = 0;        int nTmp = num;        while (nTmp != 0) {            nTmp /= 2;            nCount++;        }                for (int k = 2; k <= nCount; k++) {            int nLowerBoundary = pow(2.0, k - 1);            int nUpperBoundart = nLowerBoundary*2;            for (int i = nLowerBoundary; i < nUpperBoundart && i <= num; i++)                result[i] = result[i - nLowerBoundary] + 1;                    }        return result;    }