[leetcode]--338. Counting Bits

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0    int 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count        result[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,]