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leetcode-cn/originData/count-pairs-of-nodes.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给你一个无向图，无向图由整数 <code>n</code>  ，表示图中节点的数目，和 <code>edges</code> 组成，其中 <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> 表示 <code>u<sub>i</sub></code> 和 <code>v<sub>i</sub></code><sub> </sub>之间有一条无向边。同时给你一个代表查询的整数数组 <code>queries</code> 。</p>\n\n<p>第 <code>j</code> 个查询的答案是满足如下条件的点对 <code>(a, b)</code> 的数目：</p>\n\n<ul>\n\t<li><code>a < b</code></li>\n\t<li><code>cnt</code> 是与 <code>a</code> <strong>或者 </strong><code>b</code> 相连的边的数目，且 <code>cnt</code> <strong>严格大于 </strong><code>queries[j]</code> 。</li>\n</ul>\n\n<p>请你返回一个数组 <code>answers</code> ，其中 <code>answers.length == queries.length</code> 且 <code>answers[j]</code> 是第 <code>j</code> 个查询的答案。</p>\n\n<p>请注意，图中可能会有 <strong>重复边</strong> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://pic.leetcode-cn.com/1614828447-GMnLVg-image.png\" style=\"width: 310px; height: 278px;\" />\n<pre>\n<b>输入：</b>n = 4, edges = [[1,2],[2,4],[1,3],[2,3],[2,1]], queries = [2,3]\n<b>输出：</b>[6,5]\n<b>解释：</b>每个点对中，与至少一个点相连的边的数目如上图所示。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>n = 5, edges = [[1,5],[1,5],[3,4],[2,5],[1,3],[5,1],[2,3],[2,5]], queries = [1,2,3,4,5]\n<b>输出：</b>[10,10,9,8,6]\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 <= n <= 2 * 10<sup>4</sup></code></li>\n\t<li><code>1 <= edges.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= u<sub>i</sub>, v<sub>i</sub> <= n</code></li>\n\t<li><code>u<sub>i </sub>!= v<sub>i</sub></code></li>\n\t<li><code>1 <= queries.length <= 20</code></li>\n\t<li><code>0 <= queries[j] < edges.length</code></li>\n</ul>\n",
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@@ -149,7 +149,7 @@
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             "hints": [
                 "We want to count pairs (x,y) such that degree[x] + degree[y] - occurrences(x,y) > k",
                 "Think about iterating on x, and counting the number of valid y to pair with x.",