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.

class Solution {
    public int[] countBits(int num) {
        int[] res = new int[num + 1];
        res[0] = 0;
        for(int i = 1; i <= num; i++){
            int k = i;
            int t = 0;
            while(k != 0){
                if((k & 1) == 1) t++;
                k = k >> 1;
            }
            res[i] = t;
        }
        return res;
    }
}

位操作，注意 k & 1，因为&的优先级很低，所以需要括号。

class Solution {
    public int[] countBits(int num) {
        int[] f = new int[num + 1];
        for (int i=1; i<=num; i++) f[i] = f[i >> 1] + (i & 1);
        return f;
    }
}

这个方法有点屌，dp

dp[ i ] == 最后一位&1 + dp[ i >> 1]。