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update: 添加问题“3583.统计特殊三元组”的代码和题解 (#1211)
* 3583: WA.cpp (#1210) 0 - re compute * 3583+word: AC.cpp+en (#1210) cpp - AC,95.35%,90.92% before.设备更换前 * update: 添加问题“3583.统计特殊三元组”的代码和题解 (#1211) Signed-off-by: LetMeFly666 <[email protected]> --------- Signed-off-by: LetMeFly666 <[email protected]>
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‎Codes/3583-count-special-triplets.cpp‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-12-09 18:38:13
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-12-09 18:43:53
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*/
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#if defined(_WIN32) || defined(__APPLE__)
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#include "_[1,2]toVector.h"
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#endif
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typedef long long ll;
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const long long MOD = 1e9 + 7;
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class Solution {
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public:
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int specialTriplets(vector<int>& nums) {
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ll ans = 0;
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unordered_map<int, int> total, now;
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for (int t : nums) {
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total[t]++;
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}
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for (int t : nums) {
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int finding = t * 2;
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if (now.count(finding)) {
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ll left = now[finding];
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ans = (ans + left * (total[finding] - left - (t == 0))) % MOD;
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}
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now[t]++;
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}
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return ans;
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}
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};

‎README.md‎

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|3508.设计路由器|中等|<a href="https://leetcode.cn/problems/implement-router/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/09/20/LeetCode%203508.%E8%AE%BE%E8%AE%A1%E8%B7%AF%E7%94%B1%E5%99%A8/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/151901838" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/implement-router/solutions/3787451/letmefly-3508she-ji-lu-you-qi-stltao-stl-ehkf/" target="_blank">LeetCode题解</a>|
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|3516.找到最近的人|简单|<a href="https://leetcode.cn/problems/find-closest-person/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/09/04/LeetCode%203516.%E6%89%BE%E5%88%B0%E6%9C%80%E8%BF%91%E7%9A%84%E4%BA%BA/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/151184074" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-closest-person/solutions/3772009/letmefly-3516zhao-dao-zui-jin-de-ren-ji-8rlbz/" target="_blank">LeetCode题解</a>|
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|3541.找到频率最高的元音和辅音|简单|<a href="https://leetcode.cn/problems/find-most-frequent-vowel-and-consonant/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/09/13/LeetCode%203541.%E6%89%BE%E5%88%B0%E9%A2%91%E7%8E%87%E6%9C%80%E9%AB%98%E7%9A%84%E5%85%83%E9%9F%B3%E5%92%8C%E8%BE%85%E9%9F%B3/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/151653999" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-most-frequent-vowel-and-consonant/solutions/3780744/letmefly-3541zhao-dao-pin-lu-zui-gao-de-yv5vs/" target="_blank">LeetCode题解</a>|
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|3583.统计特殊三元组|中等|<a href="https://leetcode.cn/problems/count-special-triplets/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/12/09/LeetCode%203583.%E7%BB%9F%E8%AE%A1%E7%89%B9%E6%AE%8A%E4%B8%89%E5%85%83%E7%BB%84/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/155754955" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-special-triplets/solutions/3853381/letmefly-3583tong-ji-te-shu-san-yuan-zu-cdnx3/" target="_blank">LeetCode题解</a>|
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|剑指Offer0047.礼物的最大价值|简单|<a href="https://leetcode.cn/problems/li-wu-de-zui-da-jie-zhi-lcof/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/03/08/LeetCode%20%E5%89%91%E6%8C%87%20Offer%2047.%20%E7%A4%BC%E7%89%A9%E7%9A%84%E6%9C%80%E5%A4%A7%E4%BB%B7%E5%80%BC/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/129408765" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/li-wu-de-zui-da-jie-zhi-lcof/solutions/2155672/letmefly-jian-zhi-offer-47li-wu-de-zui-d-rekb/" target="_blank">LeetCode题解</a>|
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|剑指OfferII0041.滑动窗口的平均值|简单|<a href="https://leetcode.cn/problems/qIsx9U/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/07/16/LeetCode%20%E5%89%91%E6%8C%87%20Offer%20II%200041.%20%E6%BB%91%E5%8A%A8%E7%AA%97%E5%8F%A3%E7%9A%84%E5%B9%B3%E5%9D%87%E5%80%BC/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/125819216" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/qIsx9U/solution/by-tisfy-30mq/" target="_blank">LeetCode题解</a>|
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|剑指OfferII0091.粉刷房子|中等|<a href="https://leetcode.cn/problems/JEj789/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/06/25/LeetCode%20%E5%89%91%E6%8C%87%20Offer%20II%200091.%20%E7%B2%89%E5%88%B7%E6%88%BF%E5%AD%90/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/125456885" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/JEj789/solution/letmefly-jian-zhi-offer-ii-091fen-shua-f-3olz/" target="_blank">LeetCode题解</a>|

‎Solutions/LeetCode 3583.统计特殊三元组.md‎

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---
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title: 3583.统计特殊三元组：哈希表计数
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date: 2025-12-09 23:51:33
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tags: [题解, LeetCode, 中等, 数组, 哈希表, map, 计数]
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categories: [题解, LeetCode]
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---
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# 【LetMeFly】3583.统计特殊三元组：哈希表计数
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力扣题目链接：[https://leetcode.cn/problems/count-special-triplets/](https://leetcode.cn/problems/count-special-triplets/)
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<p>给你一个整数数组 <code>nums</code>。</p>
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<p><strong>特殊三元组 </strong>定义为满足以下条件的下标三元组 <code>(i, j, k)</code>：</p>
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<ul>
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<li><code>0 &lt;= i &lt; j &lt; k &lt; n</code>，其中 <code>n = nums.length</code></li>
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<li><code>nums[i] == nums[j] * 2</code></li>
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<li><code>nums[k] == nums[j] * 2</code></li>
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</ul>
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<p>返回数组中&nbsp;<strong>特殊三元组&nbsp;</strong>的总数。</p>
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<p>由于答案可能非常大，请返回结果对 <code>10<sup>9</sup> + 7</code> 取余数后的值。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">nums = [6,3,6]</span></p>
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<p><strong>输出：</strong> <span class="example-io">1</span></p>
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<p><strong>解释：</strong></p>
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<p>唯一的特殊三元组是 <code>(i, j, k) = (0, 1, 2)</code>，其中：</p>
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<ul>
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<li><code>nums[0] = 6</code>, <code>nums[1] = 3</code>, <code>nums[2] = 6</code></li>
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<li><code>nums[0] = nums[1] * 2 = 3 * 2 = 6</code></li>
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<li><code>nums[2] = nums[1] * 2 = 3 * 2 = 6</code></li>
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</ul>
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</div>
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<p><strong class="example">示例 2：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">nums = [0,1,0,0]</span></p>
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<p><strong>输出：</strong> <span class="example-io">1</span></p>
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<p><strong>解释：</strong></p>
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<p>唯一的特殊三元组是 <code>(i, j, k) = (0, 2, 3)</code>，其中：</p>
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<ul>
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<li><code>nums[0] = 0</code>, <code>nums[2] = 0</code>, <code>nums[3] = 0</code></li>
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<li><code>nums[0] = nums[2] * 2 = 0 * 2 = 0</code></li>
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<li><code>nums[3] = nums[2] * 2 = 0 * 2 = 0</code></li>
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</ul>
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</div>
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<p><strong class="example">示例 3：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">nums = [8,4,2,8,4]</span></p>
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<p><strong>输出：</strong> <span class="example-io">2</span></p>
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<p><strong>解释：</strong></p>
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<p>共有两个特殊三元组：</p>
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<ul>
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<li><code>(i, j, k) = (0, 1, 3)</code>
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<ul>
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<li><code>nums[0] = 8</code>, <code>nums[1] = 4</code>, <code>nums[3] = 8</code></li>
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<li><code>nums[0] = nums[1] * 2 = 4 * 2 = 8</code></li>
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<li><code>nums[3] = nums[1] * 2 = 4 * 2 = 8</code></li>
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</ul>
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</li>
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<li><code>(i, j, k) = (1, 2, 4)</code>
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<ul>
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<li><code>nums[1] = 4</code>, <code>nums[2] = 2</code>, <code>nums[4] = 4</code></li>
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<li><code>nums[1] = nums[2] * 2 = 2 * 2 = 4</code></li>
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<li><code>nums[4] = nums[2] * 2 = 2 * 2 = 4</code></li>
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</ul>
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</li>
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</ul>
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</div>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>3 &lt;= n == nums.length &lt;= 10<sup>5</sup></code></li>
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<li><code>0 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
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</ul>
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## 解题方法：哈希表计数
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使用一个哈希表，统计nums中每个数分别出现了多少次。
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接着再次遍历nums数组，再使用一个哈希表，统计到目前为止每个数分别出现了多少次。
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假设当前遍历元素为$t$，那么在此之前$2t$出现的次数乘以$2t$在此之后出现的次数($出现总次数-此前出现次数$)累加到答案中。
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注意：
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1. 特判$t=0$时
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2. 其实也可以使用64位整数并且只在最后取一次模
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### 时空复杂度分析
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+ 时间复杂度$O(len(nums))$
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+ 空间复杂度$O(len(nums))$
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### AC代码
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#### C++
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```cpp
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/*
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* @LastEditTime: 2025-12-09 18:43:53
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*/
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typedef long long ll;
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const long long MOD = 1e9 + 7;
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class Solution {
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public:
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int specialTriplets(vector<int>& nums) {
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ll ans = 0;
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unordered_map<int, int> total, now;
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for (int t : nums) {
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total[t]++;
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}
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for (int t : nums) {
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int finding = t * 2;
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if (now.count(finding)) {
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ll left = now[finding];
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ans = (ans + left * (total[finding] - left - (t == 0))) % MOD;
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}
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now[t]++;
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}
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return ans;
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}
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};
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```
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> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/155754955)和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2025/12/09/LeetCode%203583.%E7%BB%9F%E8%AE%A1%E7%89%B9%E6%AE%8A%E4%B8%89%E5%85%83%E7%BB%84/)哦~
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>
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> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)

‎Solutions/Other-English-LearningNotes-SomeWords.md‎

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|irrigate|v. 灌溉，冲洗(伤口/身体部位)|
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|heir|n. 继承人|
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|bona fide|adj. 真诚的，真实的，合法的|
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|||
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|puff|v. 吸(烟)；喷出，冒出(烟)；喘气<br/>n. 一小口(烟)，一缕，少量(烟/气)；泡芙；吹捧的文章|
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|oblige|v. (以法律/义务等)强迫；帮忙，效劳；使感激；使成为必要<br/>phrase. （feel obliged to do sth.）觉得有义务做；不得不做|
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+ 这个web要是能设计得可以闭眼(完全不睁眼)键盘控制背单词就好了。
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+ 也许可以加个AI用最近词编故事功能(返回接口中支持标注所使用单词高亮?)

‎chat.py‎

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'''
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Author: LetMeFly
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Date: 2025-11-25 11:19:58
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LastEditors: LetMeFly.xyz
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LastEditTime: 2025-12-09 23:17:49
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'''
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"""
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我想自动化LeetCode的那个项目（https://github.com/LetMeFly666/LeetCode），当前最想自动化的是每日一题。
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关于每日一题我有一个标记是从2025.4.7日开始的，这之后每天都会创建一个issue，标题格式为“[newSolution]Who can add 1 more problem of LeetCode 2873”或者部分不带“[newSolution]”的“Who can add 1 more problem of LeetCode 2873”。我想实现以下内容：
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- 将不带“[newSolution]”的相关issue加上这个标题（注意是4.7日及之后，注意是Who can add 1 more problem of xx的issue），希望可以先列举并等我确认后再执行rename
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- 统一重命名之后，将4.7日之后到今天为止的所有相关issue添加到https://github.com/users/LetMeFly666/projects/5/views/1
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- 之后每天我都想创建一个issue，内容为今天的每日一题（这可能涉及到爬虫，如果你没有较好的或现有的爬虫方案，可以先留下接口等我后续分析和实现）
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- 之后所有[newSolution]Who can add 1 more problem of LeetCode xxxx 为标题的issue全部自动添加到projects/5/views/1中
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- 如果有pr关联了这个issue，则标记状态为In Progress； 如果有issue带有“solving”标签，则标记状态为In Progress（去掉solving标签也要相应地标记回TODO状态
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额外信息：
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- 整个内容可以以任何自动化方式实现，例如归一化issueName也可以单独写一个脚本在本地运行一次
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- 4.7日之前也有类似issue，暂且不在本次统计范围内
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- 你可以使用任何方法实现，包括但不限于一个或多个workflow，或部分本地执行的gh命令或python脚本，但本地执行的只能今天执行一次，创建好后要依靠自动执行
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- 所有符合条件的issue都带有“题解”label
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是的，回答你的问题：
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①是
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②提供一个现成的 LeetCode 每日一题爬取方案
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③已创建
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④是，或者“ close: #1204”也算
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⑤全部
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额外补充：
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- 每日自动创建每日一题是UTC+8的0:10
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- issue的body内容为：“By xxx | 3583.统计特殊三元组”，其中xxx为创建脚本名称和链接，链接要链接到对应的hash地址；其中“3583.统计特殊三元组”也是个链接，要链接到这道题的地址
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现在我使用的是Windows，能否将今天运行的那些修改为python脚本呢
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请使用gh吧，我Windows系统上也有gh并且gh鉴权过了，不想再申请GITHUB_TOKEN了
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我希望你给我返回的是今天的预处理脚本SOS，不是每天运行的脚本
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这样，现在你先给我返回今天要运行的脚本，包括rename issue和添加到project
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很棒，只是顺序不太正确，毕竟我创建issue时候的顺序就不太正确，这不能怪你。
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我的每日一题正确顺序如下，请你在targets中按照此顺序排序，若不在此顺序列表中，则排序到最后。
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返回需要修改的代码。
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此外，我刚刚新增了字段Date，类型为日期，请从“ 2873 #882”开始依次填入日期字段
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这样匹配有很大的匹配失败风险，以LeetCode后面的1208为依据排序
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不是这个意思嗷，
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daily_order = [
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"[newSolution]Who can add 1 more problem of LeetCode 2873 #882",
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"[newSolution]Who can add 1 more problem of LeetCode 2874 #883",
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...
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]
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以daily_order中的顺序为依据将targets排序，但targets和daily_order中元素相同的依据为：LeetCode后的数字相同
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"""

‎r2.py‎

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