338. Counting Bits

最新推荐文章于 2022-05-28 13:29:42 发布

原创最新推荐文章于 2022-05-28 13:29:42 发布 · 97 阅读

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本文探讨了如何在给定的非负整数范围内，计算每个数字二进制表示中1的数量，并将其作为数组返回。提供了高效的算法实现，避免使用内置函数。

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 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.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/counting-bits
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

把数字分为0-3,4-7,8-16，发现每个段的值，都是前面的数字加一。

class Solution {
    public int[] countBits(int num) {
        int []res = new int[num + 1];
        int base = 1;
        int k = 1;
        while(k <= num){
            for(int j = 0; j < base && k <= num; j++)
                res[k++] = res[j] + 1;
            base <<= 1;
        }
        return res;
    }
}