【leetcode】338 .Counting Bits

原題

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.
Credits:
Special thanks to @ syedee for adding this problem and creating all test cases.

解析

給一個數字，計算從0到該數字的二進制1的個數
不容許使用內置函數

思路

個人解法。。使用了內置函數
最優解：由於*2能夠表示爲<<2,即一個數的兩倍和它的一半的1的個數和這個數相同；也即，若一個數除以2能夠除盡，則1的個數就是n/2的1的個數，若除不盡，則爲n/2+1

個人解法

public int[] countBits(int num) {
        int[] result = new int[num + 1];
        for (int i = 0; i <= num; i++) {
            result[i] = Integer.bitCount(i);
        }
        return result;
    }

最優解

public int[] countBitsOptimize(int num) {
        int[] result = new int[num + 1];
        for (int i = 1; i <= num; i++) {
            result[i] = result[i >> 1] + (i & 1);
        }
        return result;
    }