Given a non negative integer number num. For every numbersiin the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

直接的做法

class Solution {
    public int[] countBits(int num) {
        int[] res = new int[num + 1];
        int n = 0;
        int upper = 1;
        for(int i = 0; i <= num; i++) {
            res[i] = count(i);
        }
        return res;
    }

    public int count(int n) {
        int count = 0;
        for(int i = 0; i < 32; i++) {
            if(((n >> i) & 1) == 1)
                count++;
        }

        return count;
    }
}

DP

public class Solution {
    public int[] countBits(int num) {
        int[] ans = new int[num + 1];
        int i = 0, b = 1;
        // [0, b) is calculated
        while (b <= num) {
            // generate [b, 2b) or [b, num) from [0, b)
            while(i < b && i + b <= num){
                ans[i + b] = ans[i] + 1;
                ++i;
            }
            i = 0;   // reset i
            b <<= 1; // b = 2b
        }
        return ans;
    }
}