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leetcode-problemset/leetcode-cn/problem (Chinese)/计算右侧小于当前元素的个数 [count-of-smaller-numbers-after-self].html
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<p>给你一个整数数组 <code>nums</code><em> </em>,按要求返回一个新数组&nbsp;<code>counts</code><em> </em>。数组 <code>counts</code> 有该性质: <code>counts[i]</code> 的值是&nbsp; <code>nums[i]</code> 右侧小于&nbsp;<code>nums[i]</code> 的元素的数量。</p>
<p>&nbsp;</p>
<p><strong>示例 1</strong></p>
<pre>
<strong>输入:</strong>nums = [5,2,6,1]
<strong>输出:</strong><code>[2,1,1,0]
<strong>解释:</strong></code>
5 的右侧有 <strong>2 </strong>个更小的元素 (2 和 1)
2 的右侧仅有 <strong>1 </strong>个更小的元素 (1)
6 的右侧有 <strong>1 </strong>个更小的元素 (1)
1 的右侧有 <strong>0 </strong>个更小的元素
</pre>
<p><strong>示例 2</strong></p>
<pre>
<strong>输入:</strong>nums = [-1]
<strong>输出:</strong>[0]
</pre>
<p><strong>示例 3</strong></p>
<pre>
<strong>输入:</strong>nums = [-1,-1]
<strong>输出:</strong>[0,0]
</pre>
<p>&nbsp;</p>
<p><strong>提示:</strong></p>
<ul>
<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>