[LeetCode] 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:
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.

思路：

1.空间换时间，先计算出2^0 ~ 2^32 放到数组a待用。
2.将数n与数组a从大到小对比，大于某个a[i]则标记一次，并且n=n-a[i],直到对比完毕，返回标记的总次数。
3.按要求循环执行第二步，并将标记次数用数组形式输出。

Java代码：

public class Solution {
    public int[] countBits(int num) {
          int[] c = new int[num+1];
    long[] preNums=getPreNums();
    for(int i = 0; i <= num ; i++){
      c[i] = getOne(i,preNums);
      System.out.println(c[i]);
    }
   return c;
    }

     public static int getOne(int num,long[] preNums){
    long temp = num;
    int count = 0;
    for(int i = 32; temp > 0; i-- ){
      if(temp >=  preNums[i]){
        temp = temp - preNums[i];
        count++;
      }
    }
    return count;
  }

public static long[] getPreNums(){
    long[] a = new long[33];
    a[0] = 1;
    for(int i = 1; i <= 32; i++){
        a[i] = a[i-1] * 2;
    }
    return a;
}
}