Counting Bits

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问题描述（难度中等）

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:

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Input: 2
Output: [0,1,1]

Example 2:

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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.

方法一：Dynamic Programming

动态规划的思路主要来源于重复子问题，通过递推公式：f(i)=f(i>>1)+(i&1)。存在重复子问题导致递归的效率为N^2，通过动态规划时间换取空间减少时间复杂度到N。

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package P338;

import CommonUtils.ArrayUtils;

/**
 * 动态规划
 * 避免计算重复子问题
 */
class Solution {
    public int[] countBits(int num) {
        int[] counts=new int[num+1];
        for (int i = 1; i < num+1; i++) {
            //这里注意+的优先级大于&的优先级
            counts[i]=counts[i>>1]+(i&1);
        }
        return counts;
    }

    public static void main(String[] args) {
        ArrayUtils.getInstance().printIntArray(new Solution().countBits(5));
    }
}