[算法相关] 338. Counting Bits--位运算-优快云博客

本文介绍了一种解决LeetCode上题目Counting Bits的方法。该题要求计算从0到给定整数范围内每个数的二进制表示中1的个数，并返回这些个数构成的数组。文章提供了一个解决方案，并讨论了如何通过位运算提高效率。

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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.

大致题意：给定一个数，返回一个数组。这个数组为从零到这个数逐个数字中的二进制数中“1”的个数的排列。

简单的说：输入‘1’，返回‘[0,1]’,输入‘2’，返回'[0,1,1]’.就是这样：

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中的二进制转换方法，当然如果使用位运算更好了。

原题链接:https://leetcode.com/problems/counting-bits/