[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,]