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leetcode-problemset/leetcode-cn/problem (Chinese)/四数相加 II [4sum-ii].html

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<p>给你四个整数数组 <code>nums1</code><code>nums2</code><code>nums3</code><code>nums4</code> ,数组长度都是 <code>n</code> ,请你计算有多少个元组 <code>(i, j, k, l)</code> 能满足:</p>
<ul>
<li><code>0 &lt;= i, j, k, l &lt; n</code></li>
<li><code>nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0</code></li>
</ul>
<p>&nbsp;</p>
<p><strong>示例 1</strong></p>
<pre>
<strong>输入:</strong>nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]
<strong>输出:</strong>2
<strong>解释:</strong>
两个元组如下:
1. (0, 0, 0, 1) -&gt; nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -&gt; nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0
</pre>
<p><strong>示例 2</strong></p>
<pre>
<strong>输入:</strong>nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
<strong>输出:</strong>1
</pre>
<p>&nbsp;</p>
<p>&nbsp; <strong>提示:</strong></p>
<ul>
<li><code>n == nums1.length</code></li>
<li><code>n == nums2.length</code></li>
<li><code>n == nums3.length</code></li>
<li><code>n == nums4.length</code></li>
<li><code>1 &lt;= n &lt;= 200</code></li>
<li><code>-2<sup>28</sup> &lt;= nums1[i], nums2[i], nums3[i], nums4[i] &lt;= 2<sup>28</sup></code></li>
</ul>