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338. Counting Bits

338. Counting Bits

作者: HalcyonMoon | 来源:发表于2016-06-29 23:15 被阅读0次

    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(nsizeof(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.

    public class Solution {
        public int[] countBits(int num) {
            if(num == 0){
                return new int[]{0};
            }
            int dp[] = new int[num+1];
            dp[0] = 0;
            dp[1] = 1;
            for(int i = 2 ; i<=num; i++){
                if((i & 1) == 1){
                    dp[i] = dp[i>>1] + 1;
                }else{
                    dp[i] = dp[i>>1];
                }
            }
            return dp;
        }
    }
    

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