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.
遍历0~num中每个数字，求每个数字二进制中1的个数，并把结果放入数组中。
方法一：easy way。
class Solution { public: vector<int> countBits(int num) { vector<int> res; for (int i = 0; i <= num; i++) res.push_back(countBit(i)); return res; } int countBit(int num) { int res = 0; while (num) { if (num & 1) res++; num = num >> 1; } return res; } }; // 68 ms
方法二：optimization method (DP)
class Solution { public: vector<int> countBits(int num) { vector<int> res(num + 1, 0); int offset = 1; for (int i = 1; i <= num; i++) { if (offset * 2 == i) offset *= 2; res[i] = res[i - offset] + 1; } return res; } }; // 78 ms
该方法具体解法详见：How we handle this question on interview [Thinking process + DP solution]
