sort-integers-by-the-number-of-1-bits

Algorithms

Sort Integers by The Number of 1 Bits

给你一个整数数组&nbsp;<code>arr</code>&nbsp;。请你将数组中的元素按照其二进制表示中数字 1 的数目升序排序。

如果存在多个数字二进制中&nbsp;1&nbsp;的数目相同，则必须将它们按照数值大小升序排列。

请你返回排序后的数组。

&nbsp;

示例 1：

<pre>输入：arr = [0,1,2,3,4,5,6,7,8]
输出：[0,1,2,4,8,3,5,6,7]
解释：[0] 是唯一一个有 0 个 1 的数。
[1,2,4,8] 都有 1 个 1 。
[3,5,6] 有 2 个 1 。
[7] 有 3 个 1 。
按照 1 的个数排序得到的结果数组为 [0,1,2,4,8,3,5,6,7]
</pre>

示例 2：

<pre>输入：arr = [1024,512,256,128,64,32,16,8,4,2,1]
输出：[1,2,4,8,16,32,64,128,256,512,1024]
解释：数组中所有整数二进制下都只有 1 个 1 ，所以你需要按照数值大小将它们排序。
</pre>

示例 3：

<pre>输入：arr = [10000,10000]
输出：[10000,10000]
</pre>

示例 4：

<pre>输入：arr = [2,3,5,7,11,13,17,19]
输出：[2,3,5,17,7,11,13,19]
</pre>

示例 5：

<pre>输入：arr = [10,100,1000,10000]
输出：[10,100,10000,1000]
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= arr.length &lt;= 500</code></li>
	<li><code>0 &lt;= arr[i] &lt;= 10^4</code></li>
</ul>

根据数字二进制下 1 的数目排序

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sortByBits

You are given an integer array <code>arr</code>. Sort the integers in the array&nbsp;in ascending order by the number of <code>1</code>&#39;s&nbsp;in their binary representation and in case of two or more integers have the same number of <code>1</code>&#39;s you have to sort them in ascending order.

Return the array after sorting it.

&nbsp;
Example 1:

<pre>
Input: arr = [0,1,2,3,4,5,6,7,8]
Output: [0,1,2,4,8,3,5,6,7]
Explantion: [0] is the only integer with 0 bits.
[1,2,4,8] all have 1 bit.
[3,5,6] have 2 bits.
[7] has 3 bits.
The sorted array by bits is [0,1,2,4,8,3,5,6,7]
</pre>

Example 2:

<pre>
Input: arr = [1024,512,256,128,64,32,16,8,4,2,1]
Output: [1,2,4,8,16,32,64,128,256,512,1024]
Explantion: All integers have 1 bit in the binary representation, you should just sort them in ascending order.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= arr.length &lt;= 500</code></li>
	<li><code>0 &lt;= arr[i] &lt;= 104</code></li>
</ul>