本文主要是介绍[leetcode]--338. Counting Bits，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
Question 338. Counting Bits
Question: 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].中文解释：对于一个非负数的整数n，对于1~n 组成的数组，其每个数字表示的二进制数字中1的个数有多少，用数组表示输出。
Follow up:
1.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?
2.Space complexity should be O(n).
3.Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.
解决思路：迭代1~n的数组，对每个数字，利用栈的特性存储其二进制表示，然后统计每个数的二进制表示中1的个数。源代码如下：
private static int[] countBits(int num){Stack<Integer> values = new Stack<Integer>();//用栈来保存num的二进制之后的每一位数据int[] result = new int[num+1];//保存最后的结果result[0]=0;//当num为0时,结果直接是0int count=0;for(int i=1; i<=num; i++){int item = i;//取出0-num的每一项//对item取余入栈,然后再对2取整while(item>0){values.push(item%2);//对2取余的值入栈item = item/2;}//遍历栈里面的每一项,统计1的个数for(int value: values){if(value==1){count++;}}//将结果计入result中,并清零countresult[i]=count;count=0;values.clear();//栈清空}return result;
}测试代码：
public static void main(String[] args) {System.out.println("请输入num:");Scanner sc = new Scanner(System.in);String input = sc.nextLine();int num = Integer.valueOf(input);int[] result = countBits(num);System.out.print("[");for(int i: result){System.out.print(i+",");}System.out.print("]");
}运行结果:
请输入num:
10
[0,1,1,2,1,2,2,3,1,2,2,]这篇关于[leetcode]--338. Counting Bits的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
