LeetCode 338. Counting Bits(java)

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.

法一：使用内建函数，runtime为O(n*sizeof(integer))，不是最优解。

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

法二：时间最优，runtime为O(n)，通过右移一位得到之前的结果f[i>>1]，再加上最后一位是0/1即可。

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;
}

法三，runtime为O(n),通过观察发现每到2^n的时候为1，从2^n ~ 2^(n+1)，其结果为0 ~ 2^(n-1)的结果加1，因为相比较而言，就是第一个标志位多了一个1。

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;
}