338. Counting Bits（计算整数二进制表示中1的位数）

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.

Hint:
You should make use of what you have produced already.
Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
Or does the odd/even status of the number help you in calculating the number of 1s?

题目大意：给出参数num，计算0到num一共num+1个整数的二进制表示中各有多少个‘1’，将结果以数组形式返回。要求不使用时间复杂度为O(n*sizeof(integer))的蛮力算法，空间复杂度为 O(n)，且不使用相关的系统内建功能函数。

解题思路：设结果数组为res[]。
解法一：观察下表，从i=1开始的每个区间[2^0, 2^1-1],[2^1, 2^2-1],[2^2, 2^3-1]...每个区间上的res[i]都等于res[i-该区间长度]+1。

i  二进制表示 '1'  res[i]

0    0000    0    res[0]=0
-------------
1    0001    1    res[1]=res[0]+1
-------------
2    0010    1    res[2]=res[0]+1
3    0011    2    res[3]=res[1]+1
-------------
4    0100    1    res[4]=res[0]+1
5    0101    2    res[5]=res[1]+1
6    0110    2    res[6]=res[2]+1
7    0111    3    res[7]=res[3]+1
-------------
8    1000    1    ...
9    1001    2
10   1010    2
11   1011    3
12   1100    2
13   1101    3
14   1110    3
15   1111    4

那么我们可以根据这个规律得到一种解法，代码如下：（3ms，beats 41.16%）

public int[] countBits(int num) {
int[] res = new int[num + 1];
int i = 1, j, k = 2;
res[0] = 0;
while (i <= num) {
for (j = 0; i <= num && i < k; )
res[i++] = res[j++] + 1;
k <<= 1;
}
return res;
}

解法二：观察下表，发现从i=1开始，当 i 为偶数时，res[i] = res[i/2]；当
i 为奇数时，res[i] = res[i/2] + 1。这是因为i /= 2 等价于 i>>=1。当 i 为偶数时，i 的二进制表示中最低位为‘0’， i>>=1并不改变其二进制表示中‘1’的位数；当
i 为奇数时，对应的二进制最低位为‘1’， i>>=1会使 i 的二进制表示中‘1’的位数减少一位。

i  二进制表示 '1'  res[i]

0    0000    0    res[0]=0
-------------
1    0001    1    res[1]=res[0]+1
-------------
2    0010    1    res[2]=res[1]
3    0011    2    res[3]=res[1]+1
-------------
4    0100    1    res[4]=res[2]
5    0101    2    res[5]=res[2]+1
6    0110    2    res[6]=res[3]
7    0111    3    res[7]=res[3]+1
-------------
8    1000    1    ...
9    1001    2
10   1010    2
11   1011    3
12   1100    2
13   1101    3
14   1110    3
15   1111    4

解法二代码如下：（3ms，beats
41.16%）

public int[] countBits(int num) {
int[] res = new int[num + 1];
int i = 1, j, k = 2;
res[0] = 0;
while (i <= num) {
if (i % 2 == 1)
res[i] = res[i / 2] + 1;
else
res[i] = res[i / 2];
i++;
}
return res;
}

解法三：这是一种更简单的方法，规律是从i=1开始，res[i] = res[i & (i-1)] + 1。

i  二进制表示 '1'  i&(i-1)    res[i]

0    0000    0    \          res[0]=0
-------------
1    0001    1    0          res[1]=res[0]+1
-------------
2    0010    1    0          res[2]=res[0]+1
3    0011    2    2          res[3]=res[2]+1
-------------
4    0100    1    0          res[4]=res[0]+1
5    0101    2    4          res[5]=res[4]+1
6    0110    2    4          res[6]=res[4]+1
7    0111    3    6          res[7]=res[6]+1
-------------
8    1000    1    ...        ...
9    1001    2
10   1010    2
11   1011    3
12   1100    2
13   1101    3
14   1110    3
15   1111    4

解法三代码如下：(2ms，beats 87.75%)

public int[] countBits(int num) {
int[] res = new int[num + 1];
int i = 1;
res[0] = 0;
while (i <= num) {
res[i] = res[i & (i - 1)] + 1;
i++;
}
return res;
}