count-number-of-distinct-integers-after-reverse-operations

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Algorithms

You are given an array <code>nums</code> consisting of positive integers.

You have to take each integer in the array, reverse its digits, and add it to the end of the array. You should apply this operation to the original integers in <code>nums</code>.

Return the number of distinct integers in the final array.

&nbsp;
Example 1:

<pre>
Input: nums = [1,13,10,12,31]
Output: 6
Explanation: After including the reverse of each number, the resulting array is [1,13,10,12,31,1,31,1,21,13].
The reversed integers that were added to the end of the array are underlined. Note that for the integer 10, after reversing it, it becomes 01 which is just 1.
The number of distinct integers in this array is 6 (The numbers 1, 10, 12, 13, 21, and 31).</pre>

Example 2:

<pre>
Input: nums = [2,2,2]
Output: 1
Explanation: After including the reverse of each number, the resulting array is [2,2,2,2,2,2].
The number of distinct integers in this array is 1 (The number 2).
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 105</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 106</code></li>
</ul>

countDistinctIntegers

Count Number of Distinct Integers After Reverse Operations

Reverse Integer

reverse-integer

整数反转

Array

Hash Table

Math

Counting

给你一个由 正 整数组成的数组 <code>nums</code> 。

你必须取出数组中的每个整数，反转其中每个数位，并将反转后得到的数字添加到数组的末尾。这一操作只针对 <code>nums</code> 中原有的整数执行。

返回结果数组中 不同 整数的数目。

&nbsp;

示例 1：

<pre>
输入：nums = [1,13,10,12,31]
输出：6
解释：反转每个数字后，结果数组是 [1,13,10,12,31,1,31,1,21,13] 。
反转后得到的数字添加到数组的末尾并按斜体加粗表示。注意对于整数 10 ，反转之后会变成 01 ，即 1 。
数组中不同整数的数目为 6（数字 1、10、12、13、21 和 31）。</pre>

示例 2：

<pre>
输入：nums = [2,2,2]
输出：1
解释：反转每个数字后，结果数组是 [2,2,2,2,2,2] 。
数组中不同整数的数目为 1（数字 2）。
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length &lt;= 105</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 106</code></li>
</ul>