update: 添加问题“868.二进制间距”的代码和题解 (#1405)

LetMeFly666 · LetMeFly666 · commit 5857f55fa3af · 2026-02-22T10:28:54.000+08:00
Signed-off-by: LetMeFly666 &lt;Tisfy@qq.com&gt;
diff --git a/.commitTitleExtra b/.commitTitleExtra
diff --git a/.commitmsg b/.commitmsg
diff --git a/Codes/0762-prime-number-of-set-bits-in-binary-representation_20260221.cpp b/Codes/0762-prime-number-of-set-bits-in-binary-representation_20260221.cpp
@@ -2,21 +2,21 @@
  * @Author: LetMeFly
  * @Date: 2026-02-21 12:13:32
  * @LastEditors: LetMeFly.xyz
- * @LastEditTime: 2026-02-21 12:23:11
+ * @LastEditTime: 2026-02-21 12:28:09
  */
 #if defined(_WIN32) || defined(__APPLE__)
 #include "_[1,2]toVector.h"
 #endif
 
-// THIS CANNOT BE ACCEPTED
 class Solution {
 private:
     constexpr static int mask = 665772;
 public:
     int countPrimeSetBits(int left, int right) {
         int ans = 0;
         for (int i = left; i <= right; i++) {
-            if (mask && 1 << __builtin_popcount(i)) {
+            if (mask & (1 << __builtin_popcount(i))) {
+                // printf("i = %d, cnt1 = %d\n", i, __builtin_popcount(i));
                 ans++;
             }
         }
diff --git a/Codes/0762-prime-number-of-set-bits-in-binary-representation_20260221_AC.cpp b/Codes/0762-prime-number-of-set-bits-in-binary-representation_20260221_AC.cpp
diff --git a/Codes/0868-binary-gap_20260222.cpp b/Codes/0868-binary-gap_20260222.cpp
@@ -0,0 +1,26 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-02-22 09:59:17
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-02-22 10:04:00
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    int binaryGap(int n) {
+        n /= (n & -n) * 2;
+        int ans = 0, cnt = 1;
+        while (n) {
+            if (n & 1) {
+                ans = max(ans, cnt);
+                cnt = 0;
+            }
+            cnt++;
+            n >>= 1;  // Don't forget
+        }
+        return ans;
+    }
+};
diff --git a/README.md b/README.md
@@ -448,6 +448,7 @@
 |0860.柠檬水找零|简单|<a href="https://leetcode.cn/problems/lemonade-change/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/07/22/LeetCode%200860.%E6%9F%A0%E6%AA%AC%E6%B0%B4%E6%89%BE%E9%9B%B6/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131864033" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/lemonade-change/solutions/2353759/letmefly-860ning-meng-shui-zhao-ling-fu-5v6m1/" target="_blank">LeetCode题解</a>|
 |0864.获取所有钥匙的最短路径|困难|<a href="https://leetcode.cn/problems/shortest-path-to-get-all-keys/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/11/10/LeetCode%200864.%E8%8E%B7%E5%8F%96%E6%89%80%E6%9C%89%E9%92%A5%E5%8C%99%E7%9A%84%E6%9C%80%E7%9F%AD%E8%B7%AF%E5%BE%84/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127784094" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/shortest-path-to-get-all-keys/solutions/1960697/letmefly-864huo-qu-suo-you-yao-chi-de-zu-urtg/" target="_blank">LeetCode题解</a>|
 |0865.具有所有最深节点的最小子树|中等|<a href="https://leetcode.cn/problems/smallest-subtree-with-all-the-deepest-nodes/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2026/01/09/LeetCode%200865.%E5%85%B7%E6%9C%89%E6%89%80%E6%9C%89%E6%9C%80%E6%B7%B1%E8%8A%82%E7%82%B9%E7%9A%84%E6%9C%80%E5%B0%8F%E5%AD%90%E6%A0%91/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/156773047" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/smallest-subtree-with-all-the-deepest-nodes/solutions/3876835/letmefly-865ju-you-suo-you-zui-shen-jie-bg80w/" target="_blank">LeetCode题解</a>|
+|0868.二进制间距|简单|<a href="https://leetcode.cn/problems/binary-gap/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2026/02/22/LeetCode%200868.%E4%BA%8C%E8%BF%9B%E5%88%B6%E9%97%B4%E8%B7%9D/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/158283908" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/binary-gap/solutions/3906141/letmefly-868er-jin-zhi-jian-ju-wei-yun-s-jo6u/" target="_blank">LeetCode题解</a>|
 |0869.重新排序得到2的幂|中等|<a href="https://leetcode.cn/problems/reordered-power-of-2/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/08/10/LeetCode%200869.%E9%87%8D%E6%96%B0%E6%8E%92%E5%BA%8F%E5%BE%97%E5%88%B02%E7%9A%84%E5%B9%82/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/150158685" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/reordered-power-of-2/solutions/3749023/letmefly-869zhong-xin-pai-xu-de-dao-2-de-sd1j/" target="_blank">LeetCode题解</a>|
 |0870.优势洗牌|中等|<a href="https://leetcode.cn/problems/advantage-shuffle/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/10/08/LeetCode%200870.%E4%BC%98%E5%8A%BF%E6%B4%97%E7%89%8C/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127207642" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/advantage-shuffle/solution/letmefly-qu-jie-tian-ji-sai-ma-neng-ying-d9qu/" target="_blank">LeetCode题解</a>|
 |0871.最低加油次数|困难|<a href="https://leetcode.cn/problems/minimum-number-of-refueling-stops/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/07/02/LeetCode%200871.%E6%9C%80%E4%BD%8E%E5%8A%A0%E6%B2%B9%E6%AC%A1%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/125575683" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-number-of-refueling-stops/solution/letmefly-871zui-di-jia-you-ci-shu-lei-si-ucry/" target="_blank">LeetCode题解</a>|
diff --git a/Solutions/LeetCode 0868.二进制间距.md b/Solutions/LeetCode 0868.二进制间距.md
@@ -0,0 +1,108 @@
+---
+title: 868.二进制间距：位运算
+date: 2026-02-22 10:05:37
+tags: [题解, LeetCode, 简单, 位运算]
+categories: [题解, LeetCode]
+---
+
+# 【LetMeFly】868.二进制间距：位运算
+
+力扣题目链接：[https://leetcode.cn/problems/binary-gap/](https://leetcode.cn/problems/binary-gap/)
+
+<p>给定一个正整数 <code>n</code>，找到并返回 <code>n</code> 的二进制表示中两个 <strong>相邻</strong> 1 之间的<strong> 最长距离 </strong>。如果不存在两个相邻的 1，返回 <code>0</code> 。</p>
+
+<p>如果只有 <code>0</code> 将两个 <code>1</code> 分隔开（可能不存在 <code>0</code> ），则认为这两个 1 彼此 <strong>相邻</strong> 。两个 <code>1</code> 之间的距离是它们的二进制表示中位置的绝对差。例如，<code>"1001"</code> 中的两个 <code>1</code> 的距离为 3 。</p>
+
+<p>&nbsp;</p>
+
+<ul>
+</ul>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>n = 22
+<strong>输出：</strong>2
+<strong>解释：</strong>22 的二进制是 "10110" 。
+在 22 的二进制表示中，有三个 1，组成两对相邻的 1 。
+第一对相邻的 1 中，两个 1 之间的距离为 2 。
+第二对相邻的 1 中，两个 1 之间的距离为 1 。
+答案取两个距离之中最大的，也就是 2 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>n = 8
+<strong>输出：</strong>0
+<strong>解释：</strong>8 的二进制是 "1000" 。
+在 8 的二进制表示中没有相邻的两个 1，所以返回 0 。
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>n = 5
+<strong>输出：</strong>2
+<strong>解释：</strong>5 的二进制是 "101" 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+
+    
+## 解题方法：位运算
+
+以样例一的二进制`n = 10110`为例，我们只需要在`n`非零时使用一个变量`cnt`记录遇到下一个1之前一共有几位就好了，再次遇到`1`时更新答案最大值并重置`cnt`。
+
+```cpp
+while (n) {
+    if (n & 1) {
+        ans = max(ans, cnt);
+        cnt = 0;
+    }
+    cnt++;
+    n >>= 1;  // Don't forget
+}
+```
+
+但是问题在于如何得到第一个`1`的位置，很简单，可以借助[`lowbit`](https://web.letmefly.xyz/Notes/ACM/Template/lowbit.html)的思想，`n & -n`得到最低位的`1`及后面的`0`（`lowbit(10110) = 10`）令`n / lowbit(n)`即相当于抹掉了低位的所有`0`，`n / lowbit(n) >> 1`相当于抹掉了最低位的`1`及其右边的所有`0`（`10110 / lowbit(10110) >> 1 = 101`），这样就找到最低位`1`的位置了。
+
++ 时间复杂度$O(\log n)$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @LastEditTime: 2026-02-22 10:04:00
+ */
+class Solution {
+public:
+    int binaryGap(int n) {
+        n /= (n & -n) * 2;
+        int ans = 0, cnt = 1;
+        while (n) {
+            if (n & 1) {
+                ans = max(ans, cnt);
+                cnt = 0;
+            }
+            cnt++;
+            n >>= 1;  // Don't forget
+        }
+        return ans;
+    }
+};
+```
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/158283908)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/2026/02/22/LeetCode%200868.%E4%BA%8C%E8%BF%9B%E5%88%B6%E9%97%B4%E8%B7%9D/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)
diff --git a/Solutions/Other-Accumulation-SomeTips.md b/Solutions/Other-Accumulation-SomeTips.md
@@ -329,7 +329,7 @@ git config receive.denyCurrentBranch updateInstead
 
 ```bash
 gh auth login
-# 可登录多账号，之后使用gh switch USERNAME切换
+# 可登录多账号，之后使用gh auth switch -u USERNAME切换
 ```
 
 添加project权限：
diff --git a/Solutions/Other-English-LearningNotes-SomeWords.md b/Solutions/Other-English-LearningNotes-SomeWords.md
@@ -1827,6 +1827,8 @@ categories: [自用]
 |mumps|n. 腮腺炎|
 |||
 |praise|v. 赞扬<br/>n. 赞扬|
+|||
+|stipulation|n. 规定，约定，合同|
 
 + 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?)
diff --git a/Solutions/Other-Windows-RemoveExcessWindowsCredentialsGeneratedByVsCode.md b/Solutions/Other-Windows-RemoveExcessWindowsCredentialsGeneratedByVsCode.md
@@ -33,7 +33,7 @@ failed to move active token in keyring: Not enough memory resources are availabl
 
 <script src="https://web.letmefly.xyz/Links/JS/GithubGists/RmWindowsCredentials/5fc6647e2f5289a656d2441229cdd392.js"></script>
 
-<details><summary>script</summary>
+<details><summary>duplicated script(一键复制版本)</summary>
 
 
 ```powershell
@@ -115,6 +115,8 @@ Read-Host "按回车退出"
 
 删了当年留下的182个凭据后，`gh`表现正常了。
 
+附：凭据位置 - `控制面板\所有控制面板项\凭据管理器`的`Windows凭据`
+
 <center><font size="6px" face="Ink Free">The Real End, Thanks!</font></center>
 
 > 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/158262243)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/2026/02/21/Other-Windows-RemoveExcessWindowsCredentialsGeneratedByVsCode/)哦~
diff --git a/todo b/todo
@@ -1,5 +1,3 @@
-# rm xx_AC.cpp
-
 # 记录iphone快捷指令 自动化 当到达某位置时触发
 
 # move beautify.sh to github repo/gist
