# Counting Bits

Nov 23, 2017

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].

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.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0 1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1

numCount[i] = numCount[i-k] 其中k表示i只保留最高位的1时所代表的数。

vector<int> countBits(int num) {
vector<int> ret(num+1,0);
int k = 1;
for(int i = 1;i <= num;i++) {
if (i == k<<1) k<<=1;
ret[i] = ret[i - k] + 1;
}
return ret;
}

dicuss中有一些更精妙的递推式：

vector<int> countBits(int num) {
vector<int> ret(num+1, 0);
for (int i = 1; i <= num; ++i)
ret[i] = ret[i&(i-1)] + 1;
return ret;
}
LeetCodeLeetCodeDP

Diameter of Binary Tree

Hamming Distance