Question

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ numcalculate the number of 1's in their binary representation and return them as an array.

Example 1:

Input: 2 Output: [0,1,1]

Example 2:

Input: 5 Output: [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.

Difficulty:Medium

Category:

Analyze

Solution

Solution 1

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

Solution 2

对于数字 6(110)，它可以看成是 4(100) 再加一个 2(10)，因此 dp[i] = dp[i & (i-1)] + 1.

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