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].

最好把数字全都列出来然后找规律。

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

找规律就是递推，递推就是状态转移方程。这题的规律是f[i] = f[i / 2] + i % 2。

    public int[] countBits(int num) {
        int dp[] = new int[num + 1];
        dp[0] = 0;
        for (int i = 1; i <= num; i++) {
            dp[i] = dp[i / 2] + i % 2;
        }
        return dp;
    }

比较fancy的写法：

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;
}

其他思路：
http://www.cnblogs.com/grandyang/p/5294255.html

Java位运算：
http://aijuans.iteye.com/blog/1850655