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leetcode-cn/problem (Chinese)/K 个逆序对数组 [k-inverse-pairs-array].html
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<p>逆序对的定义如下：对于数组 <code>nums</code> 的第 <code>i</code> 个和第 <code>j</code> 个元素，如果满足 <code>0 &lt;= i &lt; j &lt; nums.length</code> 且&nbsp;<code>nums[i] &gt; nums[j]</code>，则其为一个逆序对；否则不是。</p>
<p>对于一个整数数组&nbsp;<code>nums</code>，<strong>逆序对</strong>是一对满足 <code>0 &lt;= i &lt; j &lt; nums.length</code> 且&nbsp;<code>nums[i] &gt; nums[j]</code>的整数对&nbsp;<code>[i, j]</code>&nbsp;。</p>
<p>给你两个整数&nbsp;<code>n</code>&nbsp;和&nbsp;<code>k</code>，找出所有包含从&nbsp;<code>1</code>&nbsp;到&nbsp;<code>n</code>&nbsp;的数字，且恰好拥有&nbsp;<code>k</code>&nbsp;个 <strong>逆序对</strong> 的不同的数组的个数。由于答案可能很大，只需要返回对 <code>10<sup>9</sup>&nbsp;+ 7</code> 取余的结果。</p>