update: 添加问题“3652.按策略买卖股票的最佳时机”的代码和题解 (#1266)

LetMeFly666 · LetMeFly666 · commit d4eefa7df2f6 · 2025-12-18T18:58:12.000+08:00
3652: AC.cpp (#1265) + word(en) + 1problem1day 1problem1day from branch 3562(#1262) Signed-off-by: LetMeFly666 <814114971@qq.com>
diff --git a/.commitmsg b/.commitmsg
@@ -0,0 +1,3 @@
+3652: AC.cpp (#1265) + word(en) + 1problem1day
+
+1problem1day from branch 3562(#1262)
diff --git a/.github/workflows/1problem1day/project-auto-add.yml b/.github/workflows/1problem1day/project-auto-add.yml
@@ -0,0 +1,66 @@
+name: Auto add 1problem1day issue to 1problem1day Project
+
+on:
+  issues:
+    types: [opened, edited]
+
+permissions:
+  contents: read
+  issues: read
+  repository-projects: write
+
+jobs:
+  add-to-project:
+    runs-on: ubuntu-latest
+    steps:
+      - name: Check title and add to project
+        env:
+          GH_TOKEN: ${{ secrets.GITHUB_TOKEN }}
+          PROJECT_ID: ${{ secrets.ONE_PROBLEM_ONE_DAY_PROJECT_ID }}
+          STATUS_FIELD_ID: ${{ secrets.ONE_PROBLEM_ONE_DAY_STATUS_FIELD_ID }}
+          DATE_FIELD_ID: ${{ secrets.ONE_PROBLEM_ONE_DAY_DATE_FIELD_ID }}
+        run: |
+          title="${{ github.event.issue.title }}"
+          issue_id="${{ github.event.issue.node_id }}"
+          created_at="${{ github.event.issue.created_at }}"
+          date_cn=$(date -d "$created_at +8 hours" +%F)
+
+          if [[ ! "$title" =~ ^\[newSolution\]Who\ can\ add\ 1\ more\ problem\ of\ LeetCode\ [0-9]+.*$ ]]; then
+            echo "Title not match, skip"
+            exit 0
+          fi
+
+          echo "Adding issue to project..."
+
+          item_id=$(gh api graphql -f query='
+            mutation($project:ID!, $content:ID!) {
+              addProjectV2ItemById(projectId:$project, contentId:$content) {
+                item { id }
+              }
+            }' -f project="$PROJECT_ID" -f content="$issue_id" --jq '.data.addProjectV2ItemById.item.id')
+
+          echo "Set Status = TODO"
+          gh api graphql -f query='
+            mutation($item:ID!, $field:ID!, $value:String!) {
+              updateProjectV2ItemFieldValue(
+                itemId:$item
+                fieldId:$field
+                value:{ singleSelectOptionId:$value }
+              ) { projectV2Item { id } }
+            }' \
+            -f item="$item_id" \
+            -f field="$STATUS_FIELD_ID" \
+            -f value="TODO"
+
+          echo "Set Date"
+          gh api graphql -f query='
+            mutation($item:ID!, $field:ID!, $value:Date!) {
+              updateProjectV2ItemFieldValue(
+                itemId:$item
+                fieldId:$field
+                value:{ date:$value }
+              ) { projectV2Item { id } }
+            }' \
+            -f item="$item_id" \
+            -f field="$DATE_FIELD_ID" \
+            -f value="${date_cn}"
diff --git a/Codes/3652-best-time-to-buy-and-sell-stock-using-strategy.cpp b/Codes/3652-best-time-to-buy-and-sell-stock-using-strategy.cpp
@@ -0,0 +1,39 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-12-18 13:17:04
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-12-18 18:42:50
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+typedef long long ll;
+class Solution {
+public:
+    ll maxProfit(vector<int>& prices, vector<int>& strategy, int k) {
+        ll ans = 0;
+        int n = prices.size();
+        for (int i = 0; i < n; i++) {
+            ans += strategy[i] * prices[i];
+        }
+
+        ll now = ans;
+        for (int i = 0; i < k / 2; i++) {
+            now += (0 - strategy[i]) * prices[i];
+        }
+        for (int i = k / 2; i < k; i++) {
+            now += (1 - strategy[i]) * prices[i];
+        }
+        ans = max(ans, now);
+
+        for (int i = 1; i + k <= n; i++) {
+            // i-1: 0->original
+            // i+k/2-1: 1->0
+            // i+k-1: original->1
+            now += (strategy[i - 1] - 0) * prices[i - 1] + (0 - 1) * prices[i + k/2 - 1] + (1 - strategy[i + k - 1]) * prices[i + k - 1];
+            ans = max(ans, now);
+        }
+        return ans;
+    }
+};
diff --git a/README.md b/README.md
@@ -1061,6 +1061,7 @@
 |3577.统计计算机解锁顺序排列数|中等|<a href="https://leetcode.cn/problems/count-the-number-of-computer-unlocking-permutations/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/12/10/LeetCode%203577.%E7%BB%9F%E8%AE%A1%E8%AE%A1%E7%AE%97%E6%9C%BA%E8%A7%A3%E9%94%81%E9%A1%BA%E5%BA%8F%E6%8E%92%E5%88%97%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/155791805" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-the-number-of-computer-unlocking-permutations/solutions/3854206/letmefly-3577tong-ji-ji-suan-ji-jie-suo-5b6b8/" target="_blank">LeetCode题解</a>|
 |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>|
 |3606.优惠券校验器|简单|<a href="https://leetcode.cn/problems/coupon-code-validator/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/12/13/LeetCode%203606.%E4%BC%98%E6%83%A0%E5%88%B8%E6%A0%A1%E9%AA%8C%E5%99%A8/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/155893803" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/coupon-code-validator/solutions/3856383/letmefly-3606you-hui-quan-xiao-yan-qi-fe-vqir/" target="_blank">LeetCode题解</a>|
+|3652.按策略买卖股票的最佳时机|中等|<a href="https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-using-strategy/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/12/18/LeetCode%203652.%E6%8C%89%E7%AD%96%E7%95%A5%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BA/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/156061117" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-using-strategy/solutions/3860245/letmefly-3652an-ce-lue-mai-mai-gu-piao-d-cnx2/" target="_blank">LeetCode题解</a>|
 |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>|
 |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>|
 |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>|
diff --git a/Solutions/LeetCode 3573.买卖股票的最佳时机V.md b/Solutions/LeetCode 3573.买卖股票的最佳时机V.md
@@ -116,7 +116,7 @@ private:
 
     ll dfs(int i, int j, int status) {
         // 0~i天 最多交易j次
-        // status: 0无1买2空头
+        // status: 0 无仓（空仓） 1 持有多头 2 持有空头
         
         int key = getKey(i, j, status);
         if (cache.count(key)) {
diff --git a/Solutions/LeetCode 3652.按策略买卖股票的最佳时机.md b/Solutions/LeetCode 3652.按策略买卖股票的最佳时机.md
@@ -0,0 +1,195 @@
+---
+title: 3652.按策略买卖股票的最佳时机：滑动窗口
+date: 2025-12-18 18:45:15
+tags: [题解, LeetCode, 中等, 数组, 前缀和, 滑动窗口]
+categories: [题解, LeetCode]
+---
+
+# 【LetMeFly】3652.按策略买卖股票的最佳时机：滑动窗口
+
+力扣题目链接：[https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-using-strategy/](https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-using-strategy/)
+
+<p>给你两个整数数组 <code>prices</code> 和 <code>strategy</code>，其中：</p>
+
+<ul>
+	<li><code>prices[i]</code> 表示第 <code>i</code> 天某股票的价格。</li>
+	<li><code>strategy[i]</code> 表示第 <code>i</code> 天的交易策略，其中：
+	<ul>
+		<li><code>-1</code> 表示买入一单位股票。</li>
+		<li><code>0</code> 表示持有股票。</li>
+		<li><code>1</code> 表示卖出一单位股票。</li>
+	</ul>
+	</li>
+</ul>
+
+<p>同时给你一个&nbsp;<strong>偶数&nbsp;</strong>整数 <code>k</code>，你可以对 <code>strategy</code> 进行&nbsp;<strong>最多一次&nbsp;</strong>修改。一次修改包括：</p>
+
+<ul>
+	<li>选择 <code>strategy</code> 中恰好 <code>k</code> 个&nbsp;<strong>连续&nbsp;</strong>元素。</li>
+	<li>将前 <code>k / 2</code> 个元素设为 <code>0</code>（持有）。</li>
+	<li>将后 <code>k / 2</code> 个元素设为 <code>1</code>（卖出）。</li>
+</ul>
+
+<p><strong>利润&nbsp;</strong>定义为所有天数中 <code>strategy[i] * prices[i]</code> 的&nbsp;<strong>总和&nbsp;</strong>。</p>
+
+<p>返回你可以获得的&nbsp;<strong>最大&nbsp;</strong>可能利润。</p>
+
+<p><strong>注意：</strong> 没有预算或股票持有数量的限制，因此所有买入和卖出操作均可行，无需考虑过去的操作。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">prices = [4,2,8], strategy = [-1,0,1], k = 2</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">10</span></p>
+
+<p><strong>解释：</strong></p>
+
+<table style="border: 1px solid black;">
+	<thead>
+		<tr>
+			<th style="border: 1px solid black;">修改</th>
+			<th style="border: 1px solid black;">策略</th>
+			<th style="border: 1px solid black;">利润计算</th>
+			<th style="border: 1px solid black;">利润</th>
+		</tr>
+	</thead>
+	<tbody>
+		<tr>
+			<td style="border: 1px solid black;">原始</td>
+			<td style="border: 1px solid black;">[-1, 0, 1]</td>
+			<td style="border: 1px solid black;">(-1 × 4) + (0 × 2) + (1 × 8) = -4 + 0 + 8</td>
+			<td style="border: 1px solid black;">4</td>
+		</tr>
+		<tr>
+			<td style="border: 1px solid black;">修改 [0, 1]</td>
+			<td style="border: 1px solid black;">[0, 1, 1]</td>
+			<td style="border: 1px solid black;">(0 × 4) + (1 × 2) + (1 × 8) = 0 + 2 + 8</td>
+			<td style="border: 1px solid black;">10</td>
+		</tr>
+		<tr>
+			<td style="border: 1px solid black;">修改 [1, 2]</td>
+			<td style="border: 1px solid black;">[-1, 0, 1]</td>
+			<td style="border: 1px solid black;">(-1 × 4) + (0 × 2) + (1 × 8) = -4 + 0 + 8</td>
+			<td style="border: 1px solid black;">4</td>
+		</tr>
+	</tbody>
+</table>
+
+<p>因此，最大可能利润是 10，通过修改子数组 <code>[0, 1]</code> 实现。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">prices = [5,4,3], strategy = [1,1,0], k = 2</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">9</span></p>
+
+<p><strong>解释：</strong></p>
+
+<div class="example-block">
+<table style="border: 1px solid black;">
+	<thead>
+		<tr>
+			<th style="border: 1px solid black;">修改</th>
+			<th style="border: 1px solid black;">策略</th>
+			<th style="border: 1px solid black;">利润计算</th>
+			<th style="border: 1px solid black;">利润</th>
+		</tr>
+	</thead>
+	<tbody>
+		<tr>
+			<td style="border: 1px solid black;">原始</td>
+			<td style="border: 1px solid black;">[1, 1, 0]</td>
+			<td style="border: 1px solid black;">(1 × 5) + (1 × 4) + (0 × 3) = 5 + 4 + 0</td>
+			<td style="border: 1px solid black;">9</td>
+		</tr>
+		<tr>
+			<td style="border: 1px solid black;">修改 [0, 1]</td>
+			<td style="border: 1px solid black;">[0, 1, 0]</td>
+			<td style="border: 1px solid black;">(0 × 5) + (1 × 4) + (0 × 3) = 0 + 4 + 0</td>
+			<td style="border: 1px solid black;">4</td>
+		</tr>
+		<tr>
+			<td style="border: 1px solid black;">修改 [1, 2]</td>
+			<td style="border: 1px solid black;">[1, 0, 1]</td>
+			<td style="border: 1px solid black;">(1 × 5) + (0 × 4) + (1 × 3) = 5 + 0 + 3</td>
+			<td style="border: 1px solid black;">8</td>
+		</tr>
+	</tbody>
+</table>
+
+<p>因此，最大可能利润是 9，无需任何修改即可达成。</p>
+</div>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>2 &lt;= prices.length == strategy.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= prices[i] &lt;= 10<sup>5</sup></code></li>
+	<li><code>-1 &lt;= strategy[i] &lt;= 1</code></li>
+	<li><code>2 &lt;= k &lt;= prices.length</code></li>
+	<li><code>k</code> 是偶数</li>
+</ul>
+
+
+    
+## 解题方法：滑动窗口
+
+既然修改范围是定长的，并且最多修改1次，那么就从前往后将每一种修改可能都试试呗。
+
+初始先计算原数组不修改时收益，再从前往后依次尝试修改区间，取收益最大的一个作为答案。
+
+如何从一个区间快速计算出下一个区间呢？变化的有3个：（变化前的）区间起点、区间中点、区间终点，把这三个位置的值更新一下就好了。
+
++ 时间复杂度$O(len(prices))$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @LastEditTime: 2025-12-18 18:42:50
+ */
+typedef long long ll;
+class Solution {
+public:
+    ll maxProfit(vector<int>& prices, vector<int>& strategy, int k) {
+        ll ans = 0;
+        int n = prices.size();
+        for (int i = 0; i < n; i++) {
+            ans += strategy[i] * prices[i];
+        }
+
+        ll now = ans;
+        for (int i = 0; i < k / 2; i++) {
+            now += (0 - strategy[i]) * prices[i];
+        }
+        for (int i = k / 2; i < k; i++) {
+            now += (1 - strategy[i]) * prices[i];
+        }
+        ans = max(ans, now);
+
+        for (int i = 1; i + k <= n; i++) {
+            // i-1: 0->original
+            // i+k/2-1: 1->0
+            // i+k-1: original->1
+            now += (strategy[i - 1] - 0) * prices[i - 1] + (0 - 1) * prices[i + k/2 - 1] + (1 - strategy[i + k - 1]) * prices[i + k - 1];
+            ans = max(ans, now);
+        }
+        return ans;
+    }
+};
+```
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/156061117)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/2025/12/18/LeetCode%203652.%E6%8C%89%E7%AD%96%E7%95%A5%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BA/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)
diff --git a/Solutions/Other-English-LearningNotes-SomeWords.md b/Solutions/Other-English-LearningNotes-SomeWords.md
@@ -1676,6 +1676,10 @@ categories: [自用]
 |||
 |bossy|adj. 好指挥人的，专横的<br/>n. 牛|
 |scarlet|adj. 猩红的，鲜红的<details><summary>例句</summary>Her lips are <font color="#28bea0">scarlet</font>.<br/>她嘴唇鲜红。</details>|
+|||
+|procession|n. 行列，列队行进，游行，（一个接着一个而来的）一队人|
+|buffalo|v. 威吓，欺骗，迷惑<br/>n. 水牛，(北美)野牛|
+|hoe|n. 锄头<br/>v. 锄地|
 
 + 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。
 + 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?)

Original file line number	Diff line number	Diff line change
`@@ -0,0 +1,3 @@`
	`1`	`+3652: AC.cpp (#1265) + word(en) + 1problem1day`
	`2`	`+`
	`3`	`+1problem1day from branch 3562(#1262)`