338. Counting Bits--位運算

338. 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:

• 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.

大致題意： 給定一個數，返回一個數組。這個數組爲從零到這個數逐個數字中的二進制數中“1”的個數的排列。

簡單的說： 輸入‘1’， 返回‘[0,1]’,輸入‘2’，返回'[0,1,1]’.就是這樣：

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

使用的是Integer中的二進制轉換方法，當然如果使用位運算更好了。

原題鏈接:https://leetcode.com/problems/counting-bits/