update: 添加问题“1758.生成交替二进制字符串的最少操作数”的代码(并更新其题解) (#1429)

* 1758: WA.cpp (#1427)

* 1758: AC.cpp (#1427) - AC,100.00%,32.67%

* update: 添加问题“1758.生成交替二进制字符串的最少操作数”的代码(并更新其题解) (#1429)

Signed-off-by: LetMeFly666 <Tisfy@qq.com>

* fix: header reg

only codex found this.

#1429 (comment)

---------

Signed-off-by: LetMeFly666 <Tisfy@qq.com>

Author: LetMeFly666

.commitmsg

@@ -21,7 +21,9 @@
             "includePath": [
                 "${workspaceFolder}/**"
             ],
-            "defines": [],
+            "defines": [
+                "_DEBUG"
+            ],
             "macFrameworkPath": [
                 "/Applications/Xcode.app/Contents/Developer/Platforms/MacOSX.platform/Developer/SDKs/MacOSX.sdk/System/Library/Frameworks"
             ],

.vscode/c_cpp_properties.json

@@ -5,6 +5,7 @@
             "label": "C/C++: g++.exe 生成活动文件",
             "command": "g++.exe",
             "args": [
+                "-D_DEBUG",
                 "-g",
                 "${file}",
                 "-o",
@@ -30,6 +31,7 @@
             "label": "C/C++: clang++ build",
             "command": "/usr/bin/clang++",
             "args": [
+                "-D_DEBUG",
                 "-fcolor-diagnostics",
                 "-fansi-escape-codes",
                 "-g",

.vscode/tasks.json

@@ -0,0 +1,21 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-03-05 22:13:51
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-03-05 22:16:57
+ */
+#ifdef _DEBUG
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    int minOperations(string s) {
+        int c0 = 0, c1 = 0;
+        for (int i = 0; i < s.size(); i++) {
+            c0 += s[i] % 2 == i % 2;  // 不用减'0'就行
+            c1 += s[i] % 2 != i % 2;
+        }
+        return min(c0, c1);
+    }
+};

Codes/1758-minimum-changes-to-make-alternating-binary-string_20260305.cpp

@@ -3,6 +3,7 @@ title: 1582.二进制矩阵中的特殊位置：模拟/记录每行每列1个数
 date: 2022-09-04 14:59:22
 tags: [题解, LeetCode, 简单, 数组, 矩阵]
 categories: [题解, LeetCode]
+index_img: https://assets.leetcode.com/uploads/2021/12/23/special1.jpg
 ---
 
 # 【LetMeFly】1582.二进制矩阵中的特殊位置：模拟/记录每行每列1个数/行有单1再看列
@@ -17,6 +18,8 @@ categories: [题解, LeetCode]
 
 <p><strong>示例 1：</strong></p>
 
+<img src="https://assets.leetcode.com/uploads/2021/12/23/special1.jpg">
+
 <pre><strong>输入：</strong>mat = [[1,0,0],
 &nbsp;           [0,0,<strong>1</strong>],
 &nbsp;           [1,0,0]]

Solutions/LeetCode 1582.二进制矩阵中的特殊位置.md

@@ -1,11 +1,11 @@
 ---
-title: 1758.生成交替二进制字符串的最少操作数
+title: 1758.生成交替二进制字符串的最少操作数：5行代码一次遍历
 date: 2022-11-29 23:00:13
 tags: [题解, LeetCode, 简单, 字符串, 模拟, 遍历]
 categories: [题解, LeetCode]
 ---
 
-# 【LetMeFly】1758.生成交替二进制字符串的最少操作数
+# 【LetMeFly】1758.生成交替二进制字符串的最少操作数：5行代码一次遍历
 
 力扣题目链接：[https://leetcode.cn/problems/minimum-changes-to-make-alternating-binary-string/](https://leetcode.cn/problems/minimum-changes-to-make-alternating-binary-string/)
 
@@ -69,6 +69,9 @@ categories: [题解, LeetCode]
 #### C++
 
 ```cpp
+/*
+ * @LastEditTime: 2022-11-29 22:56:38
+ */
 class Solution {
 public:
     int minOperations(string& s) {
@@ -82,5 +85,35 @@ public:
 };
 ```
 
+## 方法二：模拟
+
+要么`奇数`位置全`奇数` `偶数`位置全`偶数`，要么`偶数`位置全`奇数` `奇数`位置全`偶数`。
+
+两种情况两个变量一次遍历算出来取最小的那个就好了。
+
++ 时间复杂度$O(n)$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @LastEditTime: 2026-03-05 22:16:57
+ */
+class Solution {
+public:
+    int minOperations(string s) {
+        int c0 = 0, c1 = 0;
+        for (int i = 0; i < s.size(); i++) {
+            c0 += s[i] % 2 == i % 2;  // 不用减'0'就行
+            c1 += s[i] % 2 != i % 2;
+        }
+        return min(c0, c1);
+    }
+};
+```
+
 > 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)

Solutions/LeetCode 1758.生成交替二进制字符串的最少操作数.md

@@ -2,7 +2,7 @@
 Author: LetMeFly
 Date: 2022-07-03 11:21:14
 LastEditors: LetMeFly.xyz
-LastEditTime: 2026-03-01 20:15:38
+LastEditTime: 2026-03-05 22:38:46
 Command: python newSolution.py 102. 二叉树的层序遍历
 What's more: 当前仅支持数字开头的题目
 What's more: 代码结构写的很混乱 - 想单文件实现所有操作
@@ -93,7 +93,7 @@ def visit_Attribute(self, node):
         with open(toName, 'r+', encoding='utf-8') as f:
             content = f.read()
             f.seek(0)
-            header = '#if defined(_WIN32) || defined(__APPLE__)\n' +\
+            header = '#ifdef _DEBUG\n' +\
                      '#include "_[1,2]toVector.h"\n' +\
                      '#endif\n\n'
             f.write(header + content)
@@ -357,7 +357,7 @@ def removePrefix(data: str, fileType: str) -> str:# (#1399)
             new_comment = '\n'.join(kept)
             # 2. 删除紧跟的单行 #if defined(...) #include "_[1,2]toVector.h" #endif
             rest = re.sub(
-                r'^\s*#if\s+defined\([^\)]*\)(?:\s*\|\|\s*defined\([^\)]*\))*\s*\n\s*#include\s+"_\[1,2\]toVector\.h"\s*\n\s*#endif\s*\n?',
+                r'^\s*#ifdef\s+\w+\s*\n\s*#include\s+"_\[1,2\]toVector\.h"\s*\n\s*#endif\s*\n?',
                 '',
                 rest,
                 count=1

newSolution.py

@@ -0,0 +1,17 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-03-05 09:35:39
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-03-05 09:36:03
+ */
+#ifdef _DEBUG
+#include <stdio.h>
+int main() {
+    printf("1\n");
+    return 0;
+}
+#else
+int main() {
+    
+}
+#endif