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## 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 1:

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


Example 2:

Input: 5
Output: [0,1,1,2,1,2]


• 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

1. i - (i & -i) i & -i可以得到仅保留最后一位 1 的数字，例如110 & (-110)得到10
2. i & (i - 1)
class Solution {
public int[] countBits(int num) {
int[] dp = new int[num + 1];

for(int i = 1; i <= num; i++){
dp[i] = dp[i - (i & -i)] + 1;
}

return dp;
}
}

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

class Solution {
public int[] countBits(int num) {
int[] res = new int[num + 1];

for(int i = 1; i <= num; i++){
if(i % 2 == 0){
res[i] = res[i / 2];
} else {
res[i] = res[i - 1] + 1;
}
}
return res;
}
}