[LeetCode] 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:
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]