【LeetCode】338. Counting Bits

题目描述:

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.

代码实现:

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> ret(num+1,0);

        for(int i=1;i<=num;++i){
            ret[i]=ret[i>>1]+(i&1);
        }
        return ret;
    }
};
 
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