338. Counting Bits

冷不防 2022-05-23 06:25 270阅读 0赞

/*

  1. Counting Bits
    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: 

  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.

*/
/*
思路：转化成二进制字符串，去掉0，然后统计字符串长度并返回即可。
*/
class Solution {
public int[] countBits(int num) {
int[] nums = new int[num+1];
for(int i=0;i<=num;i++) {
nums[i] = Integer.valueOf( String.valueOf( Integer.toBinaryString( i ) ).replace(“0”, “”).length() );
}
return nums;
}
}

发表评论

表情：
评论列表 （有 0 条评论，270人围观）

还没有评论，来说两句吧...

相关阅读