leetcode:Counting Bits

最新推荐文章于 2019-12-07 14:42:11 发布

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本文介绍了一种计算从0到给定整数范围内所有数的二进制表示中1的数量的方法，并提供了一个Java实现示例。

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原题：
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.

题意为：给定一个非负的整数num，对于0<=i<=num的i，计算i对应二进制数中1的个数，并以数组的形式返回

Java 实现：

public class Solution {
    public int[] countBits(int num) {
        int[] result = new int[num+1];
        for (int i = 0; i <= num; i++) {
            int temp= i;
            int c=0;
            String s = Integer.toBinaryString(temp);
            for (int j = 0; j < s.length(); j++) {
                if(s.charAt(j)=='1') c++;
            }
            result[i]=c;
        }
        return result;
    }
}