lintcode/leetcode由易至难第17题：Counting Bits

Problem:

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.

Code笨方法:

public class Solution {
public int[] countBits(int num) {
int[] nums = new int[num + 1];
for(int i = 0; i <= num; i++){
int k = i;
int count = 0;
while(k != 0){
if ((k & 1) == 1){
count++;
}
k = k >> 1;
}
nums[i] = count;
}
return nums;
}
}

Code聪明方法，发现i和i/2之间存在规律:

public class Solution {
public int[] countBits(int num) {
int[] nums = new int[num + 1];
for(int i = 0; i <= num; i++){
nums[i] = nums[i / 2] + (i % 2);   //i/2可以替换成i >> 1;i%2可以替换成i&1
}
return nums;
}
}

Code聪明方法2，发现i和i-1之间存在规律：

public class Solution {
public int[] countBits(int num) {
int[] nums = new int[num + 1];
nums[0] = 0;
for(int i = 1; i <= num; i++){
nums[i] = nums[i & (i - 1)] + 1;
}
return nums;
}
}