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.

这个我用的方法就是按照传统人的思路来看1的个数的方法，+1的话，指针从最右开始，如果为0，变成1；如果为1，指针前移到为0的为止，变成1，再将之后的归0，指针再回到最右。但是这个方法时间复杂度不是很可观。

public class Counting_Bits_338 {public int[] countBits(int num) {int[] a = new int[32];int pointer = 0;int[] result = new int[num + 1];int count = 0;for (int i = 1; i <= num; i++) {if (a[pointer] == 0) {a[pointer] = 1;count++;} else {while(a[pointer]==1){pointer++;}a[pointer] = 1;count++;pointer--;while(pointer>=0){a[pointer] = 0;count--;pointer--;}pointer = 0;}result[i] = count;}return result;}public static void main(String[] args) {// TODO Auto-generated method stubCounting_Bits_338 c = new Counting_Bits_338();System.out.println(Arrays.toString(c.countBits(7)));}}

大神的方法，大神发现了规律： f[i] = f[i / 2] + i % 2.

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;}

三行解决真是太漂亮了，数学拯救人类！

还有其他的规律：

This uses the hint from the description about using ranges. Basically, the numbers in one range are equal to 1 plus all of the numbers in the ranges before it. If you write out the binary numbers, you can see that numbers 8-15 have the same pattern as 0-7 but with a 1 at the front.
My logic was to copy the previous values (starting at 0) until a power of 2 was hit (new range), at which point we just reset the t pointer back to 0 to begin the new range.

这样一看真的：[2,3]是[0,1]对应数字+1；[4,5,6,7]是[0,1,2,3]对应数字+1；[8,9,10,....,15]是[0,1,2,...,7]对应数字+1；我图片中的算法对于每个数字还要算出它的count（小于等于它的最大2指数），时间复杂度也不好，大神给出了两个指针一次遍历的算法：

public int[] countBits(int num) {    int[] ret = new int[num+1];    ret[0] = 0;    int pow = 1;    for(int i = 1, t = 0; i <= num; i++, t++) {        if(i == pow) {            pow *= 2;            t = 0;        }        ret[i] = ret[t] + 1;    }    return ret;}