338. Counting Bits

338. Counting Bits

题目：

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:

1.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?
2.Space complexity should be O(n).
3.Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

Credits:

Special thanks to @ syedee for adding this problem and creating all test cases.

代码如下：

1.一种投机取巧的方法（本题不支持）

#include <bitset>
class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> num(num + 1, 0);
        for (int i = 0; i < num + 1; i++) {
            num[i] = bitset<32>(i).count();
        }
        return num;
    }
};

2.正确解法

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> dp(num + 1, 0);
        for (int i = 1; i < num + 1; i++) {
            dp[i] = dp[i & (i - 1)] + 1;
        }
        return dp;
    }
};

解题思路：

经观察可以发现，i 的二进制表示中1的个数恰好等于i &(i-1)所得数的二进制1的个数加1：

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

且 i &(i - 1)通常可以用来判断一个数是否为2的指数，只要结果为0，就是2的指数。