LeetCode:Counting Bits

Counting Bits

   

Total Accepted: 18290 Total Submissions: 32984 Difficulty: Medium

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?

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

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 (E) Number of 1 Bits

直接贴讨论区的解答：

Question: 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.

Thinking:

1) We do not need check the input parameter, because the question has already mentioned that the number is non negative.

2) How we do this? The first idea come up with is find the pattern or rules for the result. Therefore, we can get following pattern

Index : 0 1 2 3   4 5 6 7     8 9 10 11  12 13 14 15

  num : 0 1 1 2 1 2 2 3 1 2 2 3   2   3   3  4

Do you find the pattern?

Obviously, this is overlap sub problem, and we can come up the DP solution. For now, we need find the function to implement DP.

dp[0] = 0;

dp[1] = dp[0] + 1;

dp[2] = dp[0] + 1;

dp[3] = dp[1] + 1;

dp[4] = dp[0] + 1;

dp[5] = dp[1] + 1;

dp[6] = dp[2] + 1;

dp[7] = dp[3] + 1;

dp[8] = dp[0] + 1; ...

This is the function we get, now we need find the other pattern for the function to get the general function. After we analyze the above function, we can get dp[0] = 0;

dp[1] = dp[1-1] + 1;

dp[2] = dp[2-2] + 1;

dp[3] = dp[3-2] + 1;

dp[4] = dp[4-4] + 1;

dp[5] = dp[5-4] + 1;

dp[6] = dp[6-4] + 1;

dp[7] = dp[7-4] + 1;

dp[8] = dp[8-8] + 1; ..

Obviously, we can find the pattern for above example, so now we get the general function

dp[index] = dp[index - offset] + 1;

code:

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