Leetcode no. 338

SmithZz

于 2016-06-24 14:14:13 发布

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本文链接：https://blog.youkuaiyun.com/SmithZz/article/details/51752394

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本文介绍了一种高效算法，用于计算从0到给定数值范围内每个整数的二进制表示中1的数量，并将其结果返回为数组。该算法通过巧妙利用已计算结果实现了线性时间复杂度。

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338. Counting Bits

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:

You should make use of what you have produced already.
Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
Or does the odd/even status of the number help you in calculating the number of 1s?

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