[leetcode] 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].

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:

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

核心还是要找规律。

000000

-------------------

100011

-------------------

200101

300112

-------------------

401001

501012

601102

701113

-------------------

8 10001

910012

1010102

1110113

1211002

1311013

1411103

1511114

解法一：

第一个规律：如果是i是偶数，它的bit个数跟i/2个数相同，如果是i是奇数，bit个数是i/2个数+1。ps:难找啊。。。

class Solution {public:    vector<int> countBits(int num) {        vector<int> res{0};        for(int i=1; i<=num; ++i){            if(i%2==0) res.push_back(res[i/2]);            else res.push_back(res[i/2]+1);        }        return res;            }};

解法二：

000000 i&(i-1)

-------------------

100011 0000

-------------------

200101 0000

300112 0010

-------------------

401001 0000

501012 0100

601102 0100

701113 0110

-------------------

8 10001 0000

910012 1000

1010102 1000

1110113 1010

1211002 1000

1311013 1100

1411103 1100

1511114 1110

这个规律是i的bit个数是i&(i-1)这个数的bit个数+1。

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