update: 添加问题“1888.使二进制字符串字符交替的最少反转次数”的代码和题解 (#1434)

LetMeFly666 · web-flow · commit dd8e431d5bd8 · 2026-03-08T23:52:52.000+08:00
* 1888: WA.cpp (#1431) * 1888: why.cpp (#1431) * 1888: WA.cpp (#1431) input: 001000000010 should: 4 my: 3 * 1888: why.cpp (#1431) input: 001000000010 should: 4 my: 3 * 1888: WA.cpp (#1431) input: 01001001101 should: 2 my: 5 * 1888: AC.cpp (#1431) * update: 添加问题“1888.使二进制字符串字符交替的最少反转次数”的代码和题解 (#1434) Signed-off-by: LetMeFly666 <Tisfy@qq.com> --------- Signed-off-by: LetMeFly666 <Tisfy@qq.com>
diff --git a/.commitmsg b/.commitmsg
diff --git a/Codes/1888-minimum-number-of-flips-to-make-the-binary-string-alternating.cpp b/Codes/1888-minimum-number-of-flips-to-make-the-binary-string-alternating.cpp
@@ -0,0 +1,73 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-03-07 17:46:05
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-03-08 11:41:56
+ */
+#ifdef _DEBUG
+#include "_[1,2]toVector.h"
+#endif
+
+
+/*
+01001001101
+   3次右移
+01001101010
+   ||
+01010101010
+
+相当于
+01001001101
+      ||
+01001010101
+  --
+01010101010
+_         _
+*/
+class Solution {
+public:
+    int minFlips(const string& s) {
+        int total = 0, n = s.size();
+        for (int i = 0; i < n; i++) {
+            total += s[i] % 2 == i % 2;
+        }
+        int ans = min(total, n - total);
+        if (n % 2 == 0) {
+            return ans;
+        }
+        for (int i = 0, now = 0; i < n; i++) {
+            now += s[i] % 2 == i % 2;
+            int original_front = now;
+            int original_back = total - original_front;
+            int changed_front = i + 1 - original_front;
+            int changed_back = n - i - 1 - original_back;
+            ans = min(ans, min(original_front + changed_back, changed_front + original_back));
+        }
+        return ans;
+    }
+};
+
+#if defined(_WIN32) || defined(__APPLE__)
+/*
+001000000010
+|   | | |
+101010101010
+
+001000000010
+    | | |
+001010101010
+-  右移一位
+010101010100
+   3次不可
+
+4
+*/
+int main() {
+    string s;
+    while (cin >> s) {
+        Solution sol;
+        cout << sol.minFlips(s) << endl;
+    }
+    return 0;
+}
+#endif
diff --git a/README.md b/README.md
@@ -696,6 +696,7 @@
 |1870.准时到达的列车最小时速|中等|<a href="https://leetcode.cn/problems/minimum-speed-to-arrive-on-time/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/10/02/LeetCode%201870.%E5%87%86%E6%97%B6%E5%88%B0%E8%BE%BE%E7%9A%84%E5%88%97%E8%BD%A6%E6%9C%80%E5%B0%8F%E6%97%B6%E9%80%9F/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/142680612" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-speed-to-arrive-on-time/solutions/2937033/letmefly-1870zhun-shi-dao-da-de-lie-che-liv8g/" target="_blank">LeetCode题解</a>|
 |1877.数组中最大数对和的最小值|中等|<a href="https://leetcode.cn/problems/minimize-maximum-pair-sum-in-array/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2026/01/24/LeetCode%201877.%E6%95%B0%E7%BB%84%E4%B8%AD%E6%9C%80%E5%A4%A7%E6%95%B0%E5%AF%B9%E5%92%8C%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/157333871" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimize-maximum-pair-sum-in-array/solutions/3888405/letmefly-1877shu-zu-zhong-zui-da-shu-dui-sm70/" target="_blank">LeetCode题解</a>|
 |1884.鸡蛋掉落-两枚鸡蛋|中等|<a href="https://leetcode.cn/problems/egg-drop-with-2-eggs-and-n-floors/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/10/13/LeetCode%201884.%E9%B8%A1%E8%9B%8B%E6%8E%89%E8%90%BD-%E4%B8%A4%E6%9E%9A%E9%B8%A1%E8%9B%8B/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/142906976" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/egg-drop-with-2-eggs-and-n-floors/solutions/2949710/letmefly-1884ji-dan-diao-luo-liang-mei-j-saz6/" target="_blank">LeetCode题解</a>|
+|1888.使二进制字符串字符交替的最少反转次数|中等|<a href="https://leetcode.cn/problems/minimum-number-of-flips-to-make-the-binary-string-alternating/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2026/03/07/LeetCode%201888.%E4%BD%BF%E4%BA%8C%E8%BF%9B%E5%88%B6%E5%AD%97%E7%AC%A6%E4%B8%B2%E5%AD%97%E7%AC%A6%E4%BA%A4%E6%9B%BF%E7%9A%84%E6%9C%80%E5%B0%91%E5%8F%8D%E8%BD%AC%E6%AC%A1%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/158813763" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-number-of-flips-to-make-the-binary-string-alternating/solutions/3918582/letmefly-1888shi-er-jin-zhi-zi-fu-chuan-8b104/" target="_blank">LeetCode题解</a>|
 |1895.最大的幻方|中等|<a href="https://leetcode.cn/problems/largest-magic-square/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2026/01/18/LeetCode%201895.%E6%9C%80%E5%A4%A7%E7%9A%84%E5%B9%BB%E6%96%B9/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/157104818" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/largest-magic-square/solutions/3883805/letmefly-1895zui-da-de-huan-fang-bao-li-9ln7k/" target="_blank">LeetCode题解</a>|
 |1901.寻找峰值II|中等|<a href="https://leetcode.cn/problems/find-a-peak-element-ii/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/12/19/LeetCode%201901.%E5%AF%BB%E6%89%BE%E5%B3%B0%E5%80%BCII/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135083347" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-a-peak-element-ii/solutions/2572012/letmefly-1901xun-zhao-feng-zhi-iier-fen-19tmj/" target="_blank">LeetCode题解</a>|
 |1911.最大子序列交替和|中等|<a href="https://leetcode.cn/problems/maximum-alternating-subsequence-sum/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/07/11/LeetCode%201911.%E6%9C%80%E5%A4%A7%E5%AD%90%E5%BA%8F%E5%88%97%E4%BA%A4%E6%9B%BF%E5%92%8C/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131652316" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-alternating-subsequence-sum/solutions/2339029/letmefly-1911zui-da-zi-xu-lie-jiao-ti-he-fyzq/" target="_blank">LeetCode题解</a>|
diff --git a/Solutions/LeetCode 1758.生成交替二进制字符串的最少操作数.md b/Solutions/LeetCode 1758.生成交替二进制字符串的最少操作数.md
@@ -115,5 +115,6 @@ public:
 };
 ```
 
-> 同步发文于CSDN，原创不易，转载请附上[原文链接](https://blog.letmefly.xyz/2022/11/29/LeetCode%201758.%E7%94%9F%E6%88%90%E4%BA%A4%E6%9B%BF%E4%BA%8C%E8%BF%9B%E5%88%B6%E5%AD%97%E7%AC%A6%E4%B8%B2%E7%9A%84%E6%9C%80%E5%B0%91%E6%93%8D%E4%BD%9C%E6%95%B0/)哦~
-> Tisfy：[https://letmefly.blog.csdn.net/article/details/128107132](https://letmefly.blog.csdn.net/article/details/128107132)
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/128107132)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/2022/11/29/LeetCode%201758.%E7%94%9F%E6%88%90%E4%BA%A4%E6%9B%BF%E4%BA%8C%E8%BF%9B%E5%88%B6%E5%AD%97%E7%AC%A6%E4%B8%B2%E7%9A%84%E6%9C%80%E5%B0%91%E6%93%8D%E4%BD%9C%E6%95%B0/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)
diff --git a/Solutions/LeetCode 1784.检查二进制字符串字段.md b/Solutions/LeetCode 1784.检查二进制字符串字段.md
@@ -193,6 +193,6 @@ impl Solution {
 }
 ```
 
-
-> 同步发文于CSDN，原创不易，转载请附上[原文链接](https://blog.letmefly.xyz/2022/10/03/LeetCode%201784.%E6%A3%80%E6%9F%A5%E4%BA%8C%E8%BF%9B%E5%88%B6%E5%AD%97%E7%AC%A6%E4%B8%B2%E5%AD%97%E6%AE%B5/)哦~
-> Tisfy：[https://letmefly.blog.csdn.net/article/details/127150307](https://letmefly.blog.csdn.net/article/details/127150307)
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/127150307)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/2022/10/03/LeetCode%201784.%E6%A3%80%E6%9F%A5%E4%BA%8C%E8%BF%9B%E5%88%B6%E5%AD%97%E7%AC%A6%E4%B8%B2%E5%AD%97%E6%AE%B5/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)
diff --git a/Solutions/LeetCode 1888.使二进制字符串字符交替的最少反转次数.md b/Solutions/LeetCode 1888.使二进制字符串字符交替的最少反转次数.md
@@ -0,0 +1,165 @@
+---
+title: 1888.使二进制字符串字符交替的最少反转次数：前缀和O(1)
+date: 2026-03-08 15:09:24
+tags: [题解, LeetCode, 中等, 字符串, 动态规划, 滑动窗口]
+categories: [题解, LeetCode]
+---
+
+# 【LetMeFly】1888.使二进制字符串字符交替的最少反转次数：前缀和O(1)
+
+力扣题目链接：[https://leetcode.cn/problems/minimum-number-of-flips-to-make-the-binary-string-alternating/](https://leetcode.cn/problems/minimum-number-of-flips-to-make-the-binary-string-alternating/)
+
+<p>给你一个二进制字符串 <code>s</code> 。你可以按任意顺序执行以下两种操作任意次：</p>
+
+<ul>
+	<li><strong>类型 1 ：删除</strong> 字符串 <code>s</code> 的第一个字符并将它 <strong>添加</strong> 到字符串结尾。</li>
+	<li><strong>类型 2 ：选择 </strong>字符串 <code>s</code> 中任意一个字符并将该字符 <strong>反转 </strong>，也就是如果值为 <code>'0'</code> ，则反转得到 <code>'1'</code> ，反之亦然。</li>
+</ul>
+
+<p>请你返回使 <code>s</code> 变成 <strong>交替</strong> 字符串的前提下， <strong>类型 2 </strong>的 <strong>最少</strong> 操作次数 。</p>
+
+<p>我们称一个字符串是 <strong>交替</strong> 的，需要满足任意相邻字符都不同。</p>
+
+<ul>
+	<li>比方说，字符串 <code>"010"</code> 和 <code>"1010"</code> 都是交替的，但是字符串 <code>"0100"</code> 不是。</li>
+</ul>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre><b>输入：</b>s = "111000"
+<b>输出：</b>2
+<b>解释：</b>执行第一种操作两次，得到 s = "100011" 。
+然后对第三个和第六个字符执行第二种操作，得到 s = "10<strong>1</strong>01<strong>0</strong>" 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre><b>输入：</b>s = "010"
+<b>输出：</b>0
+<strong>解释：</strong>字符串已经是交替的。
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre><b>输入：</b>s = "1110"
+<b>输出：</b>1
+<b>解释：</b>对第二个字符执行第二种操作，得到 s = "1<strong>0</strong>10" 。
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>s[i]</code> 要么是 <code>'0'</code> ，要么是 <code>'1'</code> 。</li>
+</ul>
+
+
+    
+## 解题方法：前缀和
+
+如果没有操作1（字符串轮转），那么问题就变成了[1758. 生成交替二进制字符串的最少操作数](https://leetcode.cn/problems/minimum-changes-to-make-alternating-binary-string/)。
+
+现在多了个操作1，有什么用呢？
+
+> 例如`110`，把`1`往后轮转，就变成了`101`，就不用翻转操作了。
+
+其实不难发现轮转操作只对字符串长度为奇数时候有效：
+
+> 如果字符串长度为偶数的话，最终有效的`1010`要么由`1010`变来，要么由`0101`变来。总之就是最终的01交替字符串无论是经过怎样的轮转得来的，其轮转之前就一定是01交替字符串。
+>
+> 但是奇数长度的字符串就不一样了，奇数长度的01字符串首位字符相同，所以最终有效的01字符串首位字符一定相同（如`10101`），所以就可能是原有两个相邻相同元素拆开轮转得到的（如`01101`）
+
+所以我们模拟这个“拆开的位置”就好了。对于奇数长度的字符串，先假设变成0开头0结尾：
+
+具体来说，`01101`如果全部变成`01010`需要改变3次，从前到后遍历字符串，遍历过`01`后，若从此处拆开，相当于后面本来要变成`010`的三个字符要变成`101`，后面只需要变化`len(101) - 3 = 0`次就好了。
+
+记下全部变成`010`需要反转的次数，然后从前到后遍历，记录下遍历到的位置为止变成`010..`需要反转的次数，那么后面变成`101...`需要的次数就是`len(后面字符串长度) - (原本总反转次数 - 前面已经反转次数)`
+
+讨论过变成010后，变成101同理。
+
++ 时间复杂度$O(len(s))$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @LastEditTime: 2026-03-08 11:41:56
+ */
+/*
+01001001101
+   3次右移
+01001101010
+   ||
+01010101010
+
+相当于
+01001001101
+      ||
+01001010101
+  --
+01010101010
+_         _
+*/
+class Solution {
+public:
+    int minFlips(const string& s) {
+        int total = 0, n = s.size();
+        for (int i = 0; i < n; i++) {
+            total += s[i] % 2 == i % 2;
+        }
+        int ans = min(total, n - total);
+        if (n % 2 == 0) {
+            return ans;
+        }
+        for (int i = 0, now = 0; i < n; i++) {
+            now += s[i] % 2 == i % 2;
+            int original_front = now;
+            int original_back = total - original_front;
+            int changed_front = i + 1 - original_front;
+            int changed_back = n - i - 1 - original_back;
+            ans = min(ans, min(original_front + changed_back, changed_front + original_back));
+        }
+        return ans;
+    }
+};
+
+#if defined(_WIN32) || defined(__APPLE__)
+/*
+001000000010
+|   | | |
+101010101010
+
+001000000010
+    | | |
+001010101010
+-  右移一位
+010101010100
+   3次不可
+
+4
+*/
+int main() {
+    string s;
+    while (cin >> s) {
+        Solution sol;
+        cout << sol.minFlips(s) << endl;
+    }
+    return 0;
+}
+#endif
+```
+
+## 解题方法二：滑动窗口
+
+其实也可以把字符串double，然后定长滑动窗口len(s)，求每个窗口变成010或101的最小次数。
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/158813763)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/2026/03/07/LeetCode%201888.%E4%BD%BF%E4%BA%8C%E8%BF%9B%E5%88%B6%E5%AD%97%E7%AC%A6%E4%B8%B2%E5%AD%97%E7%AC%A6%E4%BA%A4%E6%9B%BF%E7%9A%84%E6%9C%80%E5%B0%91%E5%8F%8D%E8%BD%AC%E6%AC%A1%E6%95%B0/)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
@@ -1861,6 +1861,10 @@ categories: [自用]
 |||
 |bribery|n. 贿赂，受贿|
 |bamboo shoot|phrase. 竹笋|
+|||
+|mechanics|n. 力学，结构，技巧，机械学|
+|pretentious|adj. 炫耀的，虚夸的，自命不凡的|
+|indignant|adj. 愤慨的，愤怒的|
 
 + 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
`@@ -193,6 +193,6 @@ impl Solution {`
`193`	`193`	`}`
`194`	`194`	```
`195`	`195`
`196`		`-`
`197`		`-> 同步发文于CSDN，原创不易，转载请附上[原文链接](https://blog.letmefly.xyz/2022/10/03/LeetCode%201784.%E6%A3%80%E6%9F%A5%E4%BA%8C%E8%BF%9B%E5%88%B6%E5%AD%97%E7%AC%A6%E4%B8%B2%E5%AD%97%E6%AE%B5/)哦~`
`198`		`-> Tisfy：[https://letmefly.blog.csdn.net/article/details/127150307](https://letmefly.blog.csdn.net/article/details/127150307)`
	`196`	`+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/127150307)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/2022/10/03/LeetCode%201784.%E6%A3%80%E6%9F%A5%E4%BA%8C%E8%BF%9B%E5%88%B6%E5%AD%97%E7%AC%A6%E4%B8%B2%E5%AD%97%E6%AE%B5/)哦~`
	`197`	`+>`
	`198`	`+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)`