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update: 添加问题“2483.商店的最少代价”的代码和题解 (#1285)
2483: AC.cpp (#1284) + word(en) AC,91.67%,55.95% cpp - AC,一次遍历,82.14%,98.81% Signed-off-by: LetMeFly666 <[email protected]>
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‎.commitTitleExtra‎

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‎.commitmsg‎

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2483: AC.cpp (#1284) + word(en)
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AC,91.67%,55.95%
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cpp - AC,一次遍历,82.14%,98.81%

‎Codes/2483-minimum-penalty-for-a-shop.cpp‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-12-26 18:39:14
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-12-26 18:42:18
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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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class Solution {
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public:
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int bestClosingTime(string customers) {
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int maxium = 0, ans = 0;
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for (char c : customers) {
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maxium += c == 'Y';
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}
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int thisCost = maxium;
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for (int i = 0; i < customers.size(); i++) {
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if (customers[i] == 'Y') {
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thisCost--;
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} else {
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thisCost++;
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}
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if (thisCost < maxium) {
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maxium = thisCost;
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ans = i + 1;
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}
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}
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return ans;
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}
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};

‎Codes/2483-minimum-penalty-for-a-shop_Once.cpp‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-12-26 18:39:14
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-12-26 18:46:24
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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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class Solution {
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public:
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int bestClosingTime(string customers) {
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int maxium = 0, ans = 0;
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int thisCost = 0;
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for (int i = 0; i < customers.size(); i++) {
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if (customers[i] == 'Y') {
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thisCost--;
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} else {
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thisCost++;
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}
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if (thisCost < maxium) {
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maxium = thisCost;
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ans = i + 1;
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}
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}
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return ans;
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}
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};

‎README.md‎

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|2476.二叉搜索树最近节点查询|中等|<a href="https://leetcode.cn/problems/closest-nodes-queries-in-a-binary-search-tree/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/02/24/LeetCode%202476.%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E6%9C%80%E8%BF%91%E8%8A%82%E7%82%B9%E6%9F%A5%E8%AF%A2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/136269516" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/closest-nodes-queries-in-a-binary-search-tree/solutions/2652038/letmefly-2476er-cha-sou-suo-shu-zui-jin-dbuce/" target="_blank">LeetCode题解</a>|
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|2477.到达首都的最少油耗|中等|<a href="https://leetcode.cn/problems/minimum-fuel-cost-to-report-to-the-capital/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/12/05/LeetCode%202477.%E5%88%B0%E8%BE%BE%E9%A6%96%E9%83%BD%E7%9A%84%E6%9C%80%E5%B0%91%E6%B2%B9%E8%80%97/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/134816086" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-fuel-cost-to-report-to-the-capital/solutions/2555162/letmefly-2477dao-da-shou-du-de-zui-shao-2r4od/" target="_blank">LeetCode题解</a>|
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|2481.分割圆的最少切割次数|简单|<a href="https://leetcode.cn/problems/minimum-cuts-to-divide-a-circle/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/06/17/LeetCode%202481.%E5%88%86%E5%89%B2%E5%9C%86%E7%9A%84%E6%9C%80%E5%B0%91%E5%88%87%E5%89%B2%E6%AC%A1%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131257304" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-cuts-to-divide-a-circle/solutions/2311882/letmefly-2481fen-ge-yuan-de-zui-shao-qie-w27z/" target="_blank">LeetCode题解</a>|
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|2483.商店的最少代价|中等|<a href="https://leetcode.cn/problems/minimum-penalty-for-a-shop/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/12/26/LeetCode%202483.%E5%95%86%E5%BA%97%E7%9A%84%E6%9C%80%E5%B0%91%E4%BB%A3%E4%BB%B7/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/156312519" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-penalty-for-a-shop/solutions/3866581/letmefly-2483shang-dian-de-zui-shao-dai-06zzq/" target="_blank">LeetCode题解</a>|
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|2485.找出中枢整数|简单|<a href="https://leetcode.cn/problems/find-the-pivot-integer/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/06/26/LeetCode%202485.%E6%89%BE%E5%87%BA%E4%B8%AD%E6%9E%A2%E6%95%B4%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131391141" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-the-pivot-integer/solutions/2320680/letmefly-2485zhao-chu-zhong-shu-zheng-sh-monv/" target="_blank">LeetCode题解</a>|
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|2487.从链表中移除节点|中等|<a href="https://leetcode.cn/problems/remove-nodes-from-linked-list/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/01/03/LeetCode%202487.%E4%BB%8E%E9%93%BE%E8%A1%A8%E4%B8%AD%E7%A7%BB%E9%99%A4%E8%8A%82%E7%82%B9/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135357617" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/remove-nodes-from-linked-list/solutions/2588878/letmefly-2487cong-lian-biao-zhong-yi-chu-c4dy/" target="_blank">LeetCode题解</a>|
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|2490.回环句|简单|<a href="https://leetcode.cn/problems/circular-sentence/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/06/30/LeetCode%202490.%E5%9B%9E%E7%8E%AF%E5%8F%A5/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131468453" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/circular-sentence/solutions/2325193/letmefly-2490hui-huan-ju-pan-duan-kong-g-z56k/" target="_blank">LeetCode题解</a>|

‎Solutions/LeetCode 2483.商店的最少代价.md‎

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---
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title: 2483.商店的最少代价：两次遍历 -> 一次遍历
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date: 2025-12-26 18:45:20
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tags: [题解, LeetCode, 中等, 字符串, 遍历]
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categories: [题解, LeetCode]
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---
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# 【LetMeFly】2483.商店的最少代价：两次遍历 -> 一次遍历
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力扣题目链接：[https://leetcode.cn/problems/minimum-penalty-for-a-shop/](https://leetcode.cn/problems/minimum-penalty-for-a-shop/)
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<p>给你一个顾客访问商店的日志，用一个下标从 <strong>0</strong>&nbsp;开始且只包含字符&nbsp;<code>'N'</code> 和&nbsp;<code>'Y'</code>&nbsp;的字符串&nbsp;<code>customers</code>&nbsp;表示：</p>
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<ul>
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<li>如果第&nbsp;<code>i</code>&nbsp;个字符是&nbsp;<code>'Y'</code>&nbsp;，它表示第&nbsp;<code>i</code>&nbsp;小时有顾客到达。</li>
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<li>如果第&nbsp;<code>i</code>&nbsp;个字符是&nbsp;<code>'N'</code>&nbsp;，它表示第 <code>i</code>&nbsp;小时没有顾客到达。</li>
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</ul>
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<p>如果商店在第&nbsp;<code>j</code>&nbsp;小时关门（<code>0 &lt;= j &lt;= n</code>），代价按如下方式计算：</p>
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<ul>
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<li>在开门期间，如果某一个小时没有顾客到达，代价增加 <code>1</code>&nbsp;。</li>
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<li>在关门期间，如果某一个小时有顾客到达，代价增加&nbsp;<code>1</code>&nbsp;。</li>
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</ul>
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<p>请你返回在确保代价 <strong>最小</strong>&nbsp;的前提下，商店的&nbsp;<strong>最早</strong>&nbsp;关门时间。</p>
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<p>注意，商店在第 <code>j</code>&nbsp;小时关门表示在第 <code>j</code> 小时以及之后商店处于关门状态。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<b>输入：</b>customers = "YYNY"
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<b>输出：</b>2
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<b>解释：</b>
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- 第 0 小时关门，总共 1+1+0+1 = 3 代价。
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- 第 1 小时关门，总共 0+1+0+1 = 2 代价。
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- 第 2 小时关门，总共 0+0+0+1 = 1 代价。
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- 第 3 小时关门，总共 0+0+1+1 = 2 代价。
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- 第 4 小时关门，总共 0+0+1+0 = 1 代价。
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在第 2 或第 4 小时关门代价都最小。由于第 2 小时更早，所以最优关门时间是 2 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<b>输入：</b>customers = "NNNNN"
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<b>输出：</b>0
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<b>解释：</b>最优关门时间是 0 ，因为自始至终没有顾客到达。</pre>
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<p><strong>示例 3：</strong></p>
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<pre>
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<b>输入：</b>customers = "YYYY"
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<b>输出：</b>4
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<b>解释：</b>最优关门时间是 4 ，因为每一小时均有顾客到达。
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</pre>
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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>1 &lt;= customers.length &lt;= 10<sup>5</sup></code></li>
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<li><code>customers</code>&nbsp;只包含字符&nbsp;<code>'Y'</code>&nbsp;和&nbsp;<code>'N'</code>&nbsp;。</li>
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</ul>
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## 解题方法一：两次遍历
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先计算下直接不开门的话总代价是多少。
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接着遍历顾客字符串，遍历到第i天时计算这个顾客走之后关门的总代价是多少。
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取一个最小总代价对应的关门时间；如有相同则取最早的那个。
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假设当前总代价为`x`，准备改为往后一个顾客时间再关门，那么就看下一个时间是`Y`还是`N`。如果为`Y`则总代价减1否则总代价加1。
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+ 时间复杂度$O(len(customers))$
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+ 空间复杂度$O(1)$
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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: 2025-12-26 18:42:18
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*/
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class Solution {
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public:
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int bestClosingTime(string customers) {
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int maxium = 0, ans = 0;
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for (char c : customers) {
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maxium += c == 'Y';
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}
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int thisCost = maxium;
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for (int i = 0; i < customers.size(); i++) {
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if (customers[i] == 'Y') {
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thisCost--;
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} else {
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thisCost++;
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}
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if (thisCost < maxium) {
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maxium = thisCost;
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ans = i + 1;
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}
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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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## 解题方法二：一次遍历
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不难发现第一次遍历计算得到的“直接不开门”状态下的代价其实没有意义，将其视为0且后续顾客在此基础上浮动即可。
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+ 时间复杂度$O(len(customers))$
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+ 空间复杂度$O(1)$
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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: 2025-12-26 18:46:24
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*/
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class Solution {
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public:
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int bestClosingTime(string customers) {
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int maxium = 0, ans = 0;
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int thisCost = 0;
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for (int i = 0; i < customers.size(); i++) {
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if (customers[i] == 'Y') {
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thisCost--;
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} else {
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thisCost++;
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}
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if (thisCost < maxium) {
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maxium = thisCost;
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ans = i + 1;
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}
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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/156312519)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/2025/12/26/LeetCode%202483.%E5%95%86%E5%BA%97%E7%9A%84%E6%9C%80%E5%B0%91%E4%BB%A3%E4%BB%B7/)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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|pursuance|n. 遵守，进行，实行|
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|homogenous|adj. (生物)同质的，同类的|
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|sterling|n. 英镑<br/>adj. 优秀的，杰出的，纯银的|
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|herbal|adj. 药草的，香草的<br/>n. 草本植物志|
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|arbitration|n. 仲裁|
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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?)

‎createPr.py‎

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‎createPr2.py‎

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‎getPrTitle.py‎

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