[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.

Hint:

1. You should make use of what you have produced already.
2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
3. Or does the odd/even status of the number help you in calculating the number of 1s?

解题思路

要实现O(n)的时间复杂度，显然得利用之前的计算结果。我们可以根据位运算来实现，一个数中1的位数等于它右移1位的数中1的位数加上原来的数与1相与的结果。

实现代码

// Runtime: 3 mspublic class Solution {    public int[] countBits(int num) {        int[] bits = new int[num + 1];        for (int i = 1; i <= num; i++ ) {            bits[i] = bits[i / 2] + (i & 0x1);        }        return bits;    }}