[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"$f1p2kCSJiIxWoJ7Wh8hGdGy0sTRBXbiZkv6DsWLPcwR8":3,"$fJU-4tot_gC5fDkujNeoE-cGsdMy5V_KcdUXLuAnTFgw":16,"$fES3ZHdXUBdCtO8b2EX66UsLomf2axXAE7Y_6Se8yPlg":423},{"slug":4,"title":5,"description":6,"content":7,"content_html":8,"pub_date":9,"tags":10,"draft":15},"algorithm-binary-search","二分查找：永远写不对？记住这个模板","彻底搞清楚二分查找的边界问题：闭区间和左闭右开两套模板、三道经典 LeetCode 题目完整 C++ 实现，以及二分答案的进阶思路。","二分查找是每个程序员都\"会\"但总写错的算法。\n\n不是概念不懂——找中点、比大小、缩范围，谁都明白。问题在于：`left \u003C= right` 还是 `left \u003C right`？`mid + 1` 还是 `mid`？退出循环后要不要再检查一次？\n\n死记硬背不可靠。真正的解法是**理解每套模板背后的不变式**，然后推导出所有边界条件。\n\n---\n\n## 为什么总写错\n\n根本原因是**区间语义不统一**。\n\n二分查找维护一个搜索区间，每次把目标缩小到区间的一半。但\"区间\"有两种表示方法：\n- **闭区间** `[left, right]`：两端都可能是答案\n- **左闭右开** `[left, right)`：`right` 是开边界，不在搜索范围内\n\n这两种表示导致循环条件、`mid` 更新方式、终止后的处理全都不一样。混用就出 bug。\n\n**选一套，理解它，用到底。**\n\n---\n\n## 模板一：闭区间 `[left, right]`\n\n```cpp\nint binarySearch(vector\u003Cint>& nums, int target) {\n    int left = 0, right = nums.size() - 1;  \u002F\u002F 闭区间 [left, right]\n\n    while (left \u003C= right) {  \u002F\u002F 注意：\u003C=，因为 left==right 时区间 [left,left] 仍有一个元素\n        int mid = left + (right - left) \u002F 2;  \u002F\u002F 防止 (left+right) 溢出\n\n        if (nums[mid] == target) {\n            return mid;\n        } else if (nums[mid] \u003C target) {\n            left = mid + 1;   \u002F\u002F target 在右半部分，[mid+1, right]\n        } else {\n            right = mid - 1;  \u002F\u002F target 在左半部分，[left, mid-1]\n        }\n    }\n\n    return -1;  \u002F\u002F 未找到\n}\n```\n\n**关键推导**：\n- `while (left \u003C= right)`：当 `left > right` 时区间为空，退出\n- `left = mid + 1`：已经确认 `nums[mid] \u003C target`，`mid` 不可能是答案，所以直接排除\n- `right = mid - 1`：同理，`mid` 不可能是答案\n\n适合：**确定性查找**（找到就返回，找不到返回 -1）\n\n---\n\n## 模板二：左闭右开 `[left, right)`\n\n```cpp\nint binarySearch(vector\u003Cint>& nums, int target) {\n    int left = 0, right = nums.size();  \u002F\u002F 左闭右开 [left, right)\n\n    while (left \u003C right) {  \u002F\u002F 注意：\u003C，因为 left==right 时区间为空\n        int mid = left + (right - left) \u002F 2;\n\n        if (nums[mid] == target) {\n            return mid;\n        } else if (nums[mid] \u003C target) {\n            left = mid + 1;  \u002F\u002F target 在 [mid+1, right)\n        } else {\n            right = mid;     \u002F\u002F target 在 [left, mid)，注意不是 mid-1\n        }\n    }\n\n    return -1;\n}\n```\n\n**关键推导**：\n- `right = nums.size()`（而不是 `size - 1`），因为右端是开区间\n- `while (left \u003C right)`：`left == right` 时区间为空\n- `right = mid`（不是 `mid - 1`）：因为右端是开区间，`mid` 本来就不在搜索范围内\n\n这个模板在**寻找边界**时更自然，退出循环后 `left` 就是插入位置。\n\n---\n\n## 例题 1：标准二分（LeetCode 704）\n\n> 给定升序整数数组 nums 和目标值 target，返回 target 的下标，不存在返回 -1。\n\n直接套闭区间模板：\n\n```cpp\nclass Solution {\npublic:\n    int search(vector\u003Cint>& nums, int target) {\n        int left = 0, right = (int)nums.size() - 1;\n\n        while (left \u003C= right) {\n            int mid = left + (right - left) \u002F 2;\n            if (nums[mid] == target) return mid;\n            else if (nums[mid] \u003C target) left = mid + 1;\n            else right = mid - 1;\n        }\n\n        return -1;\n    }\n};\n```\n\n---\n\n## 例题 2：搜索插入位置（LeetCode 35）\n\n> 给定排序数组和目标值，在数组中找到目标值并返回下标。如果不存在，返回它将被按顺序插入的位置。\n\n这道题要找的是**第一个 ≥ target 的位置**，用左闭右开模板最顺手：\n\n```cpp\nclass Solution {\npublic:\n    int searchInsert(vector\u003Cint>& nums, int target) {\n        int left = 0, right = (int)nums.size();  \u002F\u002F 插入位置可以是末尾，所以 right 取 size\n\n        while (left \u003C right) {\n            int mid = left + (right - left) \u002F 2;\n            if (nums[mid] \u003C target) {\n                left = mid + 1;   \u002F\u002F target 在右边\n            } else {\n                right = mid;      \u002F\u002F nums[mid] >= target，mid 可能就是答案，不排除\n            }\n        }\n\n        \u002F\u002F 循环结束时 left == right，就是插入位置\n        return left;\n    }\n};\n```\n\n**走一遍**：`nums = [1,3,5,6], target = 5`\n\n```\nleft=0, right=4\nmid=2, nums[2]=5 >= 5, right=2\nleft=0, right=2\nmid=1, nums[1]=3 \u003C 5, left=2\nleft=2, right=2, 退出\n返回 2 ✓\n```\n\n---\n\n## 例题 3：查找第一个和最后一个位置（LeetCode 34）\n\n> 在排序数组中找到目标值的第一个和最后一个位置，不存在返回 `[-1, -1]`。\n\n这是二分的经典进阶：找**左边界**和**右边界**。\n\n```cpp\nclass Solution {\npublic:\n    \u002F\u002F 找第一个 >= target 的位置（左边界）\n    int lowerBound(vector\u003Cint>& nums, int target) {\n        int left = 0, right = (int)nums.size();\n        while (left \u003C right) {\n            int mid = left + (right - left) \u002F 2;\n            if (nums[mid] \u003C target) left = mid + 1;\n            else right = mid;  \u002F\u002F nums[mid] >= target，保留\n        }\n        return left;  \u002F\u002F 第一个 >= target 的位置\n    }\n\n    vector\u003Cint> searchRange(vector\u003Cint>& nums, int target) {\n        int first = lowerBound(nums, target);\n\n        \u002F\u002F 检查 first 是否有效\n        if (first == (int)nums.size() || nums[first] != target) {\n            return {-1, -1};\n        }\n\n        \u002F\u002F 最后一个 target 的位置 = 第一个 > target 的位置 - 1\n        int last = lowerBound(nums, target + 1) - 1;\n\n        return {first, last};\n    }\n};\n```\n\n**核心技巧**：把找右边界转化为「找 `target+1` 的左边界然后 -1」，这样只需要一个 `lowerBound` 函数。\n\n---\n\n## 二分答案：不只是找元素\n\n二分查找不只能用在数组上——**任何具有单调性的答案空间**都可以二分。\n\n思路：不是在数组中找某个值，而是把「答案」本身作为搜索对象。\n\n**例：木材切割问题**\n\n> 给定 n 根木材的长度，需要切出 k 根长度为 L 的木材。找最大可行的 L。\n\n暴力：枚举所有可能的 L（1 到 max），检查能否切出 k 根 → O(max × n)\n\n二分答案：L 越大，能切出的根数越少（单调性）→ 对 L 进行二分\n\n```cpp\n\u002F\u002F 判断给定长度 L，能切出多少根\nlong long countPieces(vector\u003Cint>& logs, long long L) {\n    long long count = 0;\n    for (int log : logs) count += log \u002F L;\n    return count;\n}\n\nint maxLength(vector\u003Cint>& logs, int k) {\n    long long left = 1, right = *max_element(logs.begin(), logs.end());\n    long long ans = 0;\n\n    while (left \u003C= right) {\n        long long mid = left + (right - left) \u002F 2;\n        if (countPieces(logs, mid) >= k) {\n            ans = mid;       \u002F\u002F mid 可行，尝试更大\n            left = mid + 1;\n        } else {\n            right = mid - 1;  \u002F\u002F mid 不可行，缩小\n        }\n    }\n\n    return (int)ans;\n}\n```\n\n**二分答案的适用判断**：\n1. 答案在某个范围内\n2. 存在单调性：答案越大（或越小），某个条件越难（或越容易）满足\n3. 有一个 O(n) 或更快的 `check(x)` 函数\n\n经典题型：最小化最大值、最大化最小值、K 个分组的最优切分。\n\n---\n\n## 两套模板对比\n\n| | 闭区间 `[l, r]` | 左闭右开 `[l, r)` |\n|--|--|--|\n| 初始化 | `right = n - 1` | `right = n` |\n| 循环条件 | `left \u003C= right` | `left \u003C right` |\n| 缩小右边界 | `right = mid - 1` | `right = mid` |\n| 退出后 | 未找到返回 -1 | `left` 即插入\u002F边界位置 |\n| 适合 | 精确查找 | 边界查找、插入位置 |\n\n**记忆口诀**：左闭右开时，右端永远是「未来的 right」，所以 `right = mid`（不减 1）；退出后 left 就是答案位置，不需要额外检查。\n\n---\n\n## 总结\n\n二分查找的边界问题有且只有一个解法：**选定区间语义，严格遵守不变式**。\n\n- 闭区间：`[left, right]` 都可能是答案，循环条件 `\u003C=`，更新时两端都要 ±1\n- 左闭右开：`right` 是边界外，循环条件 `\u003C`，更新 right 时直接赋 `mid`\n\n选一套，理解为什么，剩下的推导就会水到渠成。","\u003Cp>二分查找是每个程序员都"会"但总写错的算法。\u003C\u002Fp>\n\u003Cp>不是概念不懂——找中点、比大小、缩范围，谁都明白。问题在于：\u003Ccode>left <= right\u003C\u002Fcode> 还是 \u003Ccode>left < right\u003C\u002Fcode>？\u003Ccode>mid + 1\u003C\u002Fcode> 还是 \u003Ccode>mid\u003C\u002Fcode>？退出循环后要不要再检查一次？\u003C\u002Fp>\n\u003Cp>死记硬背不可靠。真正的解法是\u003Cstrong>理解每套模板背后的不变式\u003C\u002Fstrong>，然后推导出所有边界条件。\u003C\u002Fp>\n\u003Chr>\n\u003Ch2 id=\"为什么总写错\">为什么总写错\u003C\u002Fh2>\n\u003Cp>根本原因是\u003Cstrong>区间语义不统一\u003C\u002Fstrong>。\u003C\u002Fp>\n\u003Cp>二分查找维护一个搜索区间，每次把目标缩小到区间的一半。但"区间"有两种表示方法：\u003C\u002Fp>\n\u003Cul>\n\u003Cli>\u003Cstrong>闭区间\u003C\u002Fstrong> \u003Ccode>[left, right]\u003C\u002Fcode>：两端都可能是答案\u003C\u002Fli>\n\u003Cli>\u003Cstrong>左闭右开\u003C\u002Fstrong> \u003Ccode>[left, right)\u003C\u002Fcode>：\u003Ccode>right\u003C\u002Fcode> 是开边界，不在搜索范围内\u003C\u002Fli>\n\u003C\u002Ful>\n\u003Cp>这两种表示导致循环条件、\u003Ccode>mid\u003C\u002Fcode> 更新方式、终止后的处理全都不一样。混用就出 bug。\u003C\u002Fp>\n\u003Cp>\u003Cstrong>选一套，理解它，用到底。\u003C\u002Fstrong>\u003C\u002Fp>\n\u003Chr>\n\u003Ch2 id=\"模板一-闭区间-left-right\">模板一：闭区间 \u003Ccode>[left, right]\u003C\u002Fcode>\u003C\u002Fh2>\n\u003Cpre>\u003Ccode class=\"language-cpp\">int binarySearch(vector<int>& nums, int target) {\n    int left = 0, right = nums.size() - 1;  \u002F\u002F 闭区间 [left, right]\n\n    while (left <= right) {  \u002F\u002F 注意：<=，因为 left==right 时区间 [left,left] 仍有一个元素\n        int mid = left + (right - left) \u002F 2;  \u002F\u002F 防止 (left+right) 溢出\n\n        if (nums[mid] == target) {\n            return mid;\n        } else if (nums[mid] < target) {\n            left = mid + 1;   \u002F\u002F target 在右半部分，[mid+1, right]\n        } else {\n            right = mid - 1;  \u002F\u002F target 在左半部分，[left, mid-1]\n        }\n    }\n\n    return -1;  \u002F\u002F 未找到\n}\n\u003C\u002Fcode>\u003C\u002Fpre>\n\u003Cp>\u003Cstrong>关键推导\u003C\u002Fstrong>：\u003C\u002Fp>\n\u003Cul>\n\u003Cli>\u003Ccode>while (left <= right)\u003C\u002Fcode>：当 \u003Ccode>left > right\u003C\u002Fcode> 时区间为空，退出\u003C\u002Fli>\n\u003Cli>\u003Ccode>left = mid + 1\u003C\u002Fcode>：已经确认 \u003Ccode>nums[mid] < target\u003C\u002Fcode>，\u003Ccode>mid\u003C\u002Fcode> 不可能是答案，所以直接排除\u003C\u002Fli>\n\u003Cli>\u003Ccode>right = mid - 1\u003C\u002Fcode>：同理，\u003Ccode>mid\u003C\u002Fcode> 不可能是答案\u003C\u002Fli>\n\u003C\u002Ful>\n\u003Cp>适合：\u003Cstrong>确定性查找\u003C\u002Fstrong>（找到就返回，找不到返回 -1）\u003C\u002Fp>\n\u003Chr>\n\u003Ch2 id=\"模板二-左闭右开-left-right\">模板二：左闭右开 \u003Ccode>[left, right)\u003C\u002Fcode>\u003C\u002Fh2>\n\u003Cpre>\u003Ccode class=\"language-cpp\">int binarySearch(vector<int>& nums, int target) {\n    int left = 0, right = nums.size();  \u002F\u002F 左闭右开 [left, right)\n\n    while (left < right) {  \u002F\u002F 注意：<，因为 left==right 时区间为空\n        int mid = left + (right - left) \u002F 2;\n\n        if (nums[mid] == target) {\n            return mid;\n        } else if (nums[mid] < target) {\n            left = mid + 1;  \u002F\u002F target 在 [mid+1, right)\n        } else {\n            right = mid;     \u002F\u002F target 在 [left, mid)，注意不是 mid-1\n        }\n    }\n\n    return -1;\n}\n\u003C\u002Fcode>\u003C\u002Fpre>\n\u003Cp>\u003Cstrong>关键推导\u003C\u002Fstrong>：\u003C\u002Fp>\n\u003Cul>\n\u003Cli>\u003Ccode>right = nums.size()\u003C\u002Fcode>（而不是 \u003Ccode>size - 1\u003C\u002Fcode>），因为右端是开区间\u003C\u002Fli>\n\u003Cli>\u003Ccode>while (left < right)\u003C\u002Fcode>：\u003Ccode>left == right\u003C\u002Fcode> 时区间为空\u003C\u002Fli>\n\u003Cli>\u003Ccode>right = mid\u003C\u002Fcode>（不是 \u003Ccode>mid - 1\u003C\u002Fcode>）：因为右端是开区间，\u003Ccode>mid\u003C\u002Fcode> 本来就不在搜索范围内\u003C\u002Fli>\n\u003C\u002Ful>\n\u003Cp>这个模板在\u003Cstrong>寻找边界\u003C\u002Fstrong>时更自然，退出循环后 \u003Ccode>left\u003C\u002Fcode> 就是插入位置。\u003C\u002Fp>\n\u003Chr>\n\u003Ch2 id=\"例题-1-标准二分-leetcode-704\">例题 1：标准二分（LeetCode 704）\u003C\u002Fh2>\n\u003Cblockquote>\n\u003Cp>给定升序整数数组 nums 和目标值 target，返回 target 的下标，不存在返回 -1。\u003C\u002Fp>\n\u003C\u002Fblockquote>\n\u003Cp>直接套闭区间模板：\u003C\u002Fp>\n\u003Cpre>\u003Ccode class=\"language-cpp\">class Solution {\npublic:\n    int search(vector<int>& nums, int target) {\n        int left = 0, right = (int)nums.size() - 1;\n\n        while (left <= right) {\n            int mid = left + (right - left) \u002F 2;\n            if (nums[mid] == target) return mid;\n            else if (nums[mid] < target) left = mid + 1;\n            else right = mid - 1;\n        }\n\n        return -1;\n    }\n};\n\u003C\u002Fcode>\u003C\u002Fpre>\n\u003Chr>\n\u003Ch2 id=\"例题-2-搜索插入位置-leetcode-35\">例题 2：搜索插入位置（LeetCode 35）\u003C\u002Fh2>\n\u003Cblockquote>\n\u003Cp>给定排序数组和目标值，在数组中找到目标值并返回下标。如果不存在，返回它将被按顺序插入的位置。\u003C\u002Fp>\n\u003C\u002Fblockquote>\n\u003Cp>这道题要找的是\u003Cstrong>第一个 ≥ target 的位置\u003C\u002Fstrong>，用左闭右开模板最顺手：\u003C\u002Fp>\n\u003Cpre>\u003Ccode class=\"language-cpp\">class Solution {\npublic:\n    int searchInsert(vector<int>& nums, int target) {\n        int left = 0, right = (int)nums.size();  \u002F\u002F 插入位置可以是末尾，所以 right 取 size\n\n        while (left < right) {\n            int mid = left + (right - left) \u002F 2;\n            if (nums[mid] < target) {\n                left = mid + 1;   \u002F\u002F target 在右边\n            } else {\n                right = mid;      \u002F\u002F nums[mid] >= target，mid 可能就是答案，不排除\n            }\n        }\n\n        \u002F\u002F 循环结束时 left == right，就是插入位置\n        return left;\n    }\n};\n\u003C\u002Fcode>\u003C\u002Fpre>\n\u003Cp>\u003Cstrong>走一遍\u003C\u002Fstrong>：\u003Ccode>nums = [1,3,5,6], target = 5\u003C\u002Fcode>\u003C\u002Fp>\n\u003Cpre>\u003Ccode>left=0, right=4\nmid=2, nums[2]=5 >= 5, right=2\nleft=0, right=2\nmid=1, nums[1]=3 < 5, left=2\nleft=2, right=2, 退出\n返回 2 ✓\n\u003C\u002Fcode>\u003C\u002Fpre>\n\u003Chr>\n\u003Ch2 id=\"例题-3-查找第一个和最后一个位置-leetcode-34\">例题 3：查找第一个和最后一个位置（LeetCode 34）\u003C\u002Fh2>\n\u003Cblockquote>\n\u003Cp>在排序数组中找到目标值的第一个和最后一个位置，不存在返回 \u003Ccode>[-1, -1]\u003C\u002Fcode>。\u003C\u002Fp>\n\u003C\u002Fblockquote>\n\u003Cp>这是二分的经典进阶：找\u003Cstrong>左边界\u003C\u002Fstrong>和\u003Cstrong>右边界\u003C\u002Fstrong>。\u003C\u002Fp>\n\u003Cpre>\u003Ccode class=\"language-cpp\">class Solution {\npublic:\n    \u002F\u002F 找第一个 >= target 的位置（左边界）\n    int lowerBound(vector<int>& nums, int target) {\n        int left = 0, right = (int)nums.size();\n        while (left < right) {\n            int mid = left + (right - left) \u002F 2;\n            if (nums[mid] < target) left = mid + 1;\n            else right = mid;  \u002F\u002F nums[mid] >= target，保留\n        }\n        return left;  \u002F\u002F 第一个 >= target 的位置\n    }\n\n    vector<int> searchRange(vector<int>& nums, int target) {\n        int first = lowerBound(nums, target);\n\n        \u002F\u002F 检查 first 是否有效\n        if (first == (int)nums.size() || nums[first] != target) {\n            return {-1, -1};\n        }\n\n        \u002F\u002F 最后一个 target 的位置 = 第一个 > target 的位置 - 1\n        int last = lowerBound(nums, target + 1) - 1;\n\n        return {first, last};\n    }\n};\n\u003C\u002Fcode>\u003C\u002Fpre>\n\u003Cp>\u003Cstrong>核心技巧\u003C\u002Fstrong>：把找右边界转化为「找 \u003Ccode>target+1\u003C\u002Fcode> 的左边界然后 -1」，这样只需要一个 \u003Ccode>lowerBound\u003C\u002Fcode> 函数。\u003C\u002Fp>\n\u003Chr>\n\u003Ch2 id=\"二分答案-不只是找元素\">二分答案：不只是找元素\u003C\u002Fh2>\n\u003Cp>二分查找不只能用在数组上——\u003Cstrong>任何具有单调性的答案空间\u003C\u002Fstrong>都可以二分。\u003C\u002Fp>\n\u003Cp>思路：不是在数组中找某个值，而是把「答案」本身作为搜索对象。\u003C\u002Fp>\n\u003Cp>\u003Cstrong>例：木材切割问题\u003C\u002Fstrong>\u003C\u002Fp>\n\u003Cblockquote>\n\u003Cp>给定 n 根木材的长度，需要切出 k 根长度为 L 的木材。找最大可行的 L。\u003C\u002Fp>\n\u003C\u002Fblockquote>\n\u003Cp>暴力：枚举所有可能的 L（1 到 max），检查能否切出 k 根 → O(max × n)\u003C\u002Fp>\n\u003Cp>二分答案：L 越大，能切出的根数越少（单调性）→ 对 L 进行二分\u003C\u002Fp>\n\u003Cpre>\u003Ccode class=\"language-cpp\">\u002F\u002F 判断给定长度 L，能切出多少根\nlong long countPieces(vector<int>& logs, long long L) {\n    long long count = 0;\n    for (int log : logs) count += log \u002F L;\n    return count;\n}\n\nint maxLength(vector<int>& logs, int k) {\n    long long left = 1, right = *max_element(logs.begin(), logs.end());\n    long long ans = 0;\n\n    while (left <= right) {\n        long long mid = left + (right - left) \u002F 2;\n        if (countPieces(logs, mid) >= k) {\n            ans = mid;       \u002F\u002F mid 可行，尝试更大\n            left = mid + 1;\n        } else {\n            right = mid - 1;  \u002F\u002F mid 不可行，缩小\n        }\n    }\n\n    return (int)ans;\n}\n\u003C\u002Fcode>\u003C\u002Fpre>\n\u003Cp>\u003Cstrong>二分答案的适用判断\u003C\u002Fstrong>：\u003C\u002Fp>\n\u003Col>\n\u003Cli>答案在某个范围内\u003C\u002Fli>\n\u003Cli>存在单调性：答案越大（或越小），某个条件越难（或越容易）满足\u003C\u002Fli>\n\u003Cli>有一个 O(n) 或更快的 \u003Ccode>check(x)\u003C\u002Fcode> 函数\u003C\u002Fli>\n\u003C\u002Fol>\n\u003Cp>经典题型：最小化最大值、最大化最小值、K 个分组的最优切分。\u003C\u002Fp>\n\u003Chr>\n\u003Ch2 id=\"两套模板对比\">两套模板对比\u003C\u002Fh2>\n\u003Ctable>\n\u003Cthead>\n\u003Ctr>\n\u003Cth>\u003C\u002Fth>\n\u003Cth>闭区间 \u003Ccode>[l, r]\u003C\u002Fcode>\u003C\u002Fth>\n\u003Cth>左闭右开 \u003Ccode>[l, r)\u003C\u002Fcode>\u003C\u002Fth>\n\u003C\u002Ftr>\n\u003C\u002Fthead>\n\u003Ctbody>\n\u003Ctr>\n\u003Ctd>初始化\u003C\u002Ftd>\n\u003Ctd>\u003Ccode>right = n - 1\u003C\u002Fcode>\u003C\u002Ftd>\n\u003Ctd>\u003Ccode>right = n\u003C\u002Fcode>\u003C\u002Ftd>\n\u003C\u002Ftr>\n\u003Ctr>\n\u003Ctd>循环条件\u003C\u002Ftd>\n\u003Ctd>\u003Ccode>left <= right\u003C\u002Fcode>\u003C\u002Ftd>\n\u003Ctd>\u003Ccode>left < right\u003C\u002Fcode>\u003C\u002Ftd>\n\u003C\u002Ftr>\n\u003Ctr>\n\u003Ctd>缩小右边界\u003C\u002Ftd>\n\u003Ctd>\u003Ccode>right = mid - 1\u003C\u002Fcode>\u003C\u002Ftd>\n\u003Ctd>\u003Ccode>right = mid\u003C\u002Fcode>\u003C\u002Ftd>\n\u003C\u002Ftr>\n\u003Ctr>\n\u003Ctd>退出后\u003C\u002Ftd>\n\u003Ctd>未找到返回 -1\u003C\u002Ftd>\n\u003Ctd>\u003Ccode>left\u003C\u002Fcode> 即插入\u002F边界位置\u003C\u002Ftd>\n\u003C\u002Ftr>\n\u003Ctr>\n\u003Ctd>适合\u003C\u002Ftd>\n\u003Ctd>精确查找\u003C\u002Ftd>\n\u003Ctd>边界查找、插入位置\u003C\u002Ftd>\n\u003C\u002Ftr>\n\u003C\u002Ftbody>\n\u003C\u002Ftable>\n\u003Cp>\u003Cstrong>记忆口诀\u003C\u002Fstrong>：左闭右开时，右端永远是「未来的 right」，所以 \u003Ccode>right = mid\u003C\u002Fcode>（不减 1）；退出后 left 就是答案位置，不需要额外检查。\u003C\u002Fp>\n\u003Chr>\n\u003Ch2 id=\"总结\">总结\u003C\u002Fh2>\n\u003Cp>二分查找的边界问题有且只有一个解法：\u003Cstrong>选定区间语义，严格遵守不变式\u003C\u002Fstrong>。\u003C\u002Fp>\n\u003Cul>\n\u003Cli>闭区间：\u003Ccode>[left, right]\u003C\u002Fcode> 都可能是答案，循环条件 \u003Ccode><=\u003C\u002Fcode>，更新时两端都要 ±1\u003C\u002Fli>\n\u003Cli>左闭右开：\u003Ccode>right\u003C\u002Fcode> 是边界外，循环条件 \u003Ccode><\u003C\u002Fcode>，更新 right 时直接赋 \u003Ccode>mid\u003C\u002Fcode>\u003C\u002Fli>\n\u003C\u002Ful>\n\u003Cp>选一套，理解为什么，剩下的推导就会水到渠成。\u003C\u002Fp>\n","2026-04-30",[11,12,13,14],"算法","二分查找","leetcode","cpp",false,[17,30,41,53,63,70,77,84,91,98,108,117,127,136,143,151,160,169,178,188,195,204,210,216,222,231,233,240,248,258,267,276,286,296,306,314,324,335,345,354,362,368,376,384,392,400,408,415],{"slug":18,"title":19,"description":20,"pub_date":21,"tags":22,"draft":15,"word_count":29},"ide-skills-guide","Agent Skills 完全指南：21 款第三方 Skill 深度评测与使用心得","全面评测 21 款第三方 Agent Skills，涵盖 Vue 生态、前端设计、构建工具、实用工具四大分类。从安装配置到实际使用场景，带你了解每个 Skill 的功能特点、最佳实践与使用心得。","2026-06-15",[23,24,25,26,27,28],"agent","skills","AI","效率工具","前端","Vue",4169,{"slug":31,"title":32,"description":33,"pub_date":34,"tags":35,"draft":15,"word_count":40},"linux-kernel-skeleton-struct-funcptr-container_of","Linux 内核骨架：struct、函数指针与 container_of","读懂 Linux 内核源码的三件套：巨大的 struct 组合代替继承、函数指针表实现虚派发、container_of 宏从嵌入成员找回完整对象。","2026-05-09",[36,37,38,39],"linux","kernel","C","container_of",1369,{"slug":42,"title":43,"description":44,"pub_date":45,"tags":46,"draft":15,"word_count":52},"astro-complete-guide-2025","Astro 5 深度剖析：Islands 架构原理、构建优化与 Cloudflare Workers 边缘部署","从编译器视角解析 Astro 5 的 Islands 架构实现原理，Content Layer API 的 Vite 插件机制，Server Islands 的流式渲染，以及如何在 Cloudflare Workers + D1 边缘环境下榨干性能。","2026-05-08",[47,48,49,50,51],"astro","frontend","cloudflare","performance","architecture",3663,{"slug":54,"title":55,"description":56,"pub_date":57,"tags":58,"draft":15,"word_count":62},"llm-prompt-engineering","Prompt Engineering 实战：让 LLM 真正听话的技巧","System prompt 怎么写、Few-shot 怎么设计、Chain-of-Thought 原理，以及常见失败模式和调试方法。","2026-05-03",[59,60,61],"ai","llm","工程实践",1723,{"slug":64,"title":65,"description":66,"pub_date":57,"tags":67,"draft":15,"word_count":69},"rag-system-design","RAG 系统设计：从 naive 到 production-ready","Retrieval-Augmented Generation 不只是「向量数据库 + LLM」，分块策略、召回质量、重排序、缓存才是工程核心。",[59,68,60,61],"rag",1613,{"slug":71,"title":72,"description":73,"pub_date":57,"tags":74,"draft":15,"word_count":76},"git-advanced-workflow","Git 进阶工作流：rebase、cherry-pick、bisect 的正确使用","merge 会了，但 rebase 总搞错？bisect 找 bug 提交？interactive rebase 整理历史？这篇一次说清楚。",[75,61],"git",1396,{"slug":78,"title":79,"description":80,"pub_date":57,"tags":81,"draft":15,"word_count":83},"docker-practical-guide","Docker 实战：从会用到用好","会 docker run 不够，Dockerfile 最佳实践、多阶段构建、Compose 编排、镜像瘦身才是日常真正需要的。",[82,36,61],"docker",1268,{"slug":85,"title":86,"description":87,"pub_date":57,"tags":88,"draft":15,"word_count":90},"anthropics-skills-guide","anthropics\u002Fskills：Anthropic 官方 Agent Skills 仓库解析","Anthropic 官方开源的 Agent Skills 标准仓库，127k stars，解析 SKILL.md 规范、17 个示例 skill 的设计模式，以及如何在 Claude Code \u002F Claude.ai \u002F API 中使用",[59,89,23,24],"Claude",2090,{"slug":92,"title":93,"description":94,"pub_date":57,"tags":95,"draft":15,"word_count":97},"karpathy-claude-code-guidelines","Karpathy 的 LLM 编码批评与 CLAUDE.md 最佳实践","基于 Andrej Karpathy 对 LLM 编程助手的观察，forrestchang 提炼出一个 CLAUDE.md 文件，4 条原则解决 AI 编码的典型失控问题：乱猜假设、过度设计、乱改代码、目标不清",[59,89,96,61],"Claude Code",2699,{"slug":99,"title":100,"description":101,"pub_date":57,"tags":102,"draft":15,"word_count":107},"typescript-advanced-patterns","TypeScript 高级模式：让类型系统为你工作","基础 TS 会了但类型总是 any？条件类型、映射类型、模板字面量类型、infer 关键字才是 TS 的真正威力。",[103,104,105,106],"typescript","类型系统","前端工程","高级模式",1419,{"slug":109,"title":110,"description":111,"pub_date":57,"tags":112,"draft":15,"word_count":116},"linux-performance-tuning","Linux 性能调优实战：从 top 到 perf 的完整工具链","遇到性能问题不知道从哪下手？这篇建立系统化的排查思路，从 CPU\u002F内存\u002FIO\u002F网络逐层分析。",[36,113,114,115],"性能","运维","系统编程",1524,{"slug":118,"title":119,"description":120,"pub_date":57,"tags":121,"draft":15,"word_count":126},"python-functional-programming","Python 函数式编程：map\u002Ffilter\u002Freduce 之外","Python 不是纯函数式语言，但 functools、itertools、偏函数、闭包这些工具用好了能让代码简洁一个量级。",[122,123,124,125],"python","函数式","闭包","装饰器",1867,{"slug":128,"title":129,"description":130,"pub_date":57,"tags":131,"draft":15,"word_count":135},"python-oop-guide","Python 面向对象：__init__ 之外你需要知道的","Python OOP 不只是 class + __init__，魔术方法、描述符、元类才是真正的武器。",[122,132,133,134],"OOP","面向对象","魔术方法",1792,{"slug":137,"title":138,"description":139,"pub_date":57,"tags":140,"draft":15,"word_count":142},"python-data-structures","Python 内置数据结构深度解析","list、dict、set、tuple 不只是数据容器，搞懂它们的底层实现和时间复杂度，才能写出高性能 Python。",[122,141,113,11],"数据结构",1517,{"slug":144,"title":145,"description":146,"pub_date":57,"tags":147,"draft":15,"word_count":150},"python-basics-quick-start","Python 快速上手：写给有编程基础的人","已经会其他语言，想快速掌握 Python 的语法特性和思维方式，这篇是捷径。",[122,148,149],"入门","基础",1607,{"slug":152,"title":153,"description":154,"pub_date":57,"tags":155,"draft":15,"word_count":159},"python-dataclass-pydantic","Python dataclass vs Pydantic：数据类选型指南","dataclass 是标准库的轻量选择，Pydantic v2 是带验证的重武器，什么时候用哪个，这篇说清楚。",[122,156,157,158],"dataclass","pydantic","数据验证",1323,{"slug":161,"title":162,"description":163,"pub_date":57,"tags":164,"draft":15,"word_count":168},"python-asyncio-practical","Python asyncio 实战：从回调地狱到协程优雅","asyncio 是 Python 异步编程的核心，搞懂 event loop、Task、gather 这些概念才能写出真正高效的异步代码。",[122,165,166,167],"asyncio","并发","网络编程",1258,{"slug":170,"title":171,"description":172,"pub_date":57,"tags":173,"draft":15,"word_count":177},"python-type-hints-guide","Python 类型注解完全指南：从入门到实践","Python 3.5+ 引入类型注解，配合 mypy\u002Fpyright 让 Python 也能享受静态类型检查的好处。",[122,174,175,176],"typescript-style","type-hints","工具链",1102,{"slug":179,"title":180,"description":181,"pub_date":182,"tags":183,"draft":15,"word_count":187},"pwa-install-update-button","PWA 踩坑：为什么安装按钮从来不出现","从 beforeinstallprompt 到 Service Worker waiting，把 PWA 的安装与更新提示真正做对","2026-05-02",[184,185,186],"pwa","javascript","web",1683,{"slug":189,"title":190,"description":191,"pub_date":192,"tags":193,"draft":15,"word_count":194},"openclaw-vs-hermes-agent","OpenClaw vs Hermes Agent：两个本地优先 Agent 的设计差异","OpenClaw（Novita AI）和 Hermes Agent（Nous Research）都是本地运行的个人 AI Agent，但在记忆系统、技能学习、运行环境和模型生态上走了不同的路。深入对比两种架构的核心差异。","2026-05-01",[59,23,60],1679,{"slug":196,"title":197,"description":198,"pub_date":192,"tags":199,"draft":15,"word_count":203},"cpp-random-design-patterns","C++ 设计模式实战：RAII、观察者、工厂","用现代 C++（C++17\u002F20）实现三种高频设计模式：RAII 资源管理、观察者模式事件系统、工厂模式插件架构。每种模式给出问题场景、实现代码和真实工程案例。",[14,200,201,202],"设计模式","c++17","工程",2613,{"slug":205,"title":206,"description":207,"pub_date":192,"tags":208,"draft":15,"word_count":209},"data-structures-fundamentals","数据结构基础：从数组到红黑树","系统梳理常用数据结构的核心原理、时间复杂度和适用场景。数组、链表、栈、队列、哈希表、二叉树、堆、图，每种结构附实现要点和 C++ 代码片段。",[141,11,14,149],3004,{"slug":211,"title":212,"description":213,"pub_date":9,"tags":214,"draft":15,"word_count":215},"ai-agent-what-is","什么是 AI Agent？从 LLM 到自主执行","LLM 本身是无状态问答机，Agent 是什么让它’动’起来的？本文深入解析 Agent 的四个核心能力、ReAct 框架、工具调用原理，以及主流框架横向对比。",[59,23,60],2116,{"slug":217,"title":218,"description":219,"pub_date":9,"tags":220,"draft":15,"word_count":221},"ai-agent-memory","AI Agent 的记忆系统：从上下文窗口到长期记忆","深入拆解 AI Agent 的四种记忆类型、上下文窗口压缩策略、RAG 向量检索原理，以及三种典型失败模式和工程选型建议。",[59,23,68],2052,{"slug":223,"title":224,"description":225,"pub_date":9,"tags":226,"draft":15,"word_count":230},"network-proxy-vpn-guide","代理与翻墙技术原理：从 HTTP 代理到现代协议","深入解析代理与 VPN 的本质区别，梳理从 SOCKS5 到 Shadowsocks、V2Ray\u002FXray、Hysteria2 的协议演进，以及机场订阅的技术本质。",[227,228,229],"网络","代理","协议",2148,{"slug":4,"title":5,"description":6,"pub_date":9,"tags":232,"draft":15,"word_count":150},[11,12,13,14],{"slug":234,"title":235,"description":236,"pub_date":9,"tags":237,"draft":15,"word_count":239},"algorithm-sliding-window","滑动窗口算法：从暴力到 O(n) 的思维跃迁","系统讲解滑动窗口算法的核心模板、适用题型，配合三道经典 LeetCode 题目的完整 C++ 实现，彻底理解双指针收缩思路。",[11,238,13,14],"滑动窗口",1943,{"slug":241,"title":242,"description":243,"pub_date":9,"tags":244,"draft":15,"word_count":247},"network-clash-config","Clash \u002F Mihomo 配置详解：规则、策略组与分流","深入解析 Clash\u002FMihomo 的核心配置结构，包括代理节点、策略组类型、规则优先级、DNS fake-ip 模式，以及一份实用的完整配置模板。",[227,245,228,246],"clash","配置",1292,{"slug":249,"title":250,"description":251,"pub_date":252,"tags":253,"draft":15,"word_count":257},"hid-hotplug","HID 设备热插拔检测：从 udev 到 node-hid","在 Linux 上用 node-hid + usb 库实现可靠的 USB HID 设备热插拔检测，踩坑记录","2026-04-28",[14,254,36,255,256],"hid","nodejs","electron",2039,{"slug":259,"title":260,"description":261,"pub_date":262,"tags":263,"draft":15,"word_count":266},"electron-ipc-types","Electron IPC 类型安全：从 any 到完全类型化","用 TypeScript 泛型封装 Electron IPC，彻底消灭 any，preload 契约集中管理","2026-04-25",[256,103,264,265],"ipc","vue",1446,{"slug":268,"title":269,"description":270,"pub_date":271,"tags":272,"draft":15,"word_count":275},"element-plus-popover-hide","手动关闭多个 el-popover（不用 v-model:visible）","通过 ref + Reflect.get 调用 hide() 方法手动关闭 Element Plus Popover，解释 Vue3 Proxy 导致无法直接调用实例方法的原因。","2024-10-25",[265,273,274],"element-plus","vue3",1321,{"slug":277,"title":278,"description":279,"pub_date":280,"tags":281,"draft":15,"word_count":285},"vite-vue3-ts-elementplus-pinia","用 Vite+（vp）从零搭建 Vue3 + TypeScript + Element Plus + Pinia + Vue Router","使用 Vite+ 统一工具链（vp）一条命令搭建 Vue3 全家桶，涵盖按需导入、Pinia store、路由配置，以及常见坑的解决方案。","2024-08-27",[265,282,103,273,283,284],"vite","pinia","vite-plus",1960,{"slug":287,"title":288,"description":289,"pub_date":290,"tags":291,"draft":15,"word_count":295},"cef-lnk2038-iterator-debug-level","CEF LNK2038：解决 _ITERATOR_DEBUG_LEVEL 不匹配错误","分析 CEF（Chromium Embedded Framework）集成时出现的 LNK2038 _ITERATOR_DEBUG_LEVEL 链接错误，从根本原因到解决方案的完整指南。","2024-05-07",[14,292,293,294],"CEF","Visual Studio","链接错误",1509,{"slug":297,"title":298,"description":299,"pub_date":300,"tags":301,"draft":15,"word_count":305},"npm-electron-install-fix","彻底解决 npm 安装 Electron 失败的问题","分析 npm install electron 失败的根本原因（下载二进制超时\u002F被墙），通过国内镜像（npmmirror）彻底解决，并介绍多种备选方案和常见错误排查。","2024-03-01",[256,302,303,304],"npm","前端工具链","国内镜像",1494,{"slug":307,"title":308,"description":309,"pub_date":310,"tags":311,"draft":15,"word_count":313},"git-out-of-memory","解决 git 报错：Fatal: Out of memory, malloc failed","分析 git 大仓库操作时出现 Out of memory malloc failed 的根本原因，通过调整 pack.windowMemory、http.postBuffer 和 git repack 彻底解决。","2024-01-31",[75,36,312],"工具",2244,{"slug":315,"title":316,"description":317,"pub_date":318,"tags":319,"draft":15,"word_count":323},"vmware-tools-install","在 VMware 虚拟机中安装 open-vm-tools 完整指南","详解 VMware Tools 的作用、open-vm-tools 与官方 VMware Tools 的区别，以及在 Ubuntu 虚拟机中安装并生效的完整步骤和常见问题排查。","2023-11-21",[320,36,321,322],"VMware","Ubuntu","虚拟机",2523,{"slug":325,"title":326,"description":327,"pub_date":328,"tags":329,"draft":15,"word_count":334},"load-balancing-algorithms","负载均衡算法完全指南：从轮询到一致性哈希","系统梳理静态与动态负载均衡算法，涵盖轮询、随机、权重、IP Hash、一致性 Hash、最少连接、最快响应等，并对比 Nginx、Dubbo、Spring Cloud LoadBalancer 的实现差异。","2023-11-15",[330,331,332,333],"分布式","负载均衡","Nginx","微服务",1764,{"slug":336,"title":337,"description":338,"pub_date":339,"tags":340,"draft":15,"word_count":344},"win-cw2a-ca2w","ATL 字符串转换：CW2A 与 CA2W 完全指南","详解 ATL 宏 CW2A\u002FCA2W 在 Unicode 与 ANSI 之间的字符串转换用法、头文件依赖、USES_CONVERSION 宏的作用与常见陷阱。","2023-06-09",[14,341,342,343],"windows","ATL","字符串",1665,{"slug":346,"title":347,"description":348,"pub_date":339,"tags":349,"draft":15,"word_count":353},"csharp-sendmessage-cpp","C# 通过 SendMessage 向 C++ 窗口发送消息与字符串","使用 P\u002FInvoke 调用 user32.dll 的 SendMessage，从 C# 发送自定义 WM_USER 消息及字符串指针给 C++ 原生窗口，并在 C++ 侧正确接收和转换。",[350,14,341,351,352],"C#","互操作","PInvoke",1554,{"slug":355,"title":356,"description":357,"pub_date":358,"tags":359,"draft":15,"word_count":361},"win-postmessage-vector","Windows PostMessage 跨线程传递 std::vector 指针","通过 PostMessage 在 Windows 消息队列中传递 std::vector 指针，使用 reinterpret_cast 将指针装入 LPARAM，并在接收方正确释放内存。","2023-05-26",[14,341,360],"WinAPI",1823,{"slug":363,"title":364,"description":365,"pub_date":358,"tags":366,"draft":15,"word_count":367},"exe-dll-single-package","将 EXE 和 DLL 打包成单一可执行文件","介绍两种将 exe 和依赖 dll 打包成单文件的方案：Enigma Virtual Box 和 WinRAR 自解压，适合发布 Windows 桌面程序时简化分发流程。",[341,14,312],1619,{"slug":369,"title":370,"description":371,"pub_date":358,"tags":372,"draft":15,"word_count":375},"cpp-random-mt19937","C++ 现代随机数生成：用 mt19937 彻底告别 rand()","深入讲解为什么 rand() 不够用，以及如何用 C++11 的 \u003Crandom> 库正确生成高质量随机数，涵盖 mt19937、各种分布和线程安全。",[14,373,374],"c++11","random",1549,{"slug":377,"title":378,"description":379,"pub_date":380,"tags":381,"draft":15,"word_count":383},"win-startup-registry","C++ 实现程序开机自启动：注册表方式详解","通过操作 Windows 注册表 Run 键实现程序开机自启动，包括 HKCU 与 HKLM 区别、完整封装代码、工作目录问题和 UAC 权限处理。","2022-12-26",[341,14,382],"registry",1201,{"slug":385,"title":386,"description":387,"pub_date":388,"tags":389,"draft":15,"word_count":391},"mfc-cstring-wparam","MFC 中 CString 与 WPARAM 之间的转换","详解 MFC 消息传递中 CString 无法直接强转为 WPARAM 的原因，以及两种正确的转换方案，并介绍结构体指针传递的正确姿势。","2022-11-25",[390,14,341],"mfc",1546,{"slug":393,"title":394,"description":395,"pub_date":396,"tags":397,"draft":15,"word_count":399},"duilib-static-build","正确编译 Duilib 静态库：避免 ATL 依赖和链接错误","详解如何用 DuiLib_Static.vcxproj 编译 Duilib 静态库，解决 VARIANT 未定义、Unicode 配置不匹配和 ATL 依赖等常见问题。","2022-08-24",[14,398,341,390],"duilib",2639,{"slug":401,"title":402,"description":403,"pub_date":404,"tags":405,"draft":15,"word_count":407},"mfc-dpi-adaptive","MFC 界面自适应不同分辨率","MFC 对话框程序实现控件和字体随分辨率自动缩放的完整方案，附 DPI Awareness 配置说明","2022-08-17",[390,14,341,406],"dpi",1414,{"slug":409,"title":410,"description":411,"pub_date":412,"tags":413,"draft":15,"word_count":414},"mfc-drag-window","MFC 无标题栏窗口客户区拖动：三种方法对比","MFC 对话框去掉标题栏后如何实现拖动移动窗口，三种方案完整实现与适用场景分析","2022-08-16",[390,14,341],1633,{"slug":416,"title":417,"description":418,"pub_date":419,"tags":420,"draft":15,"word_count":422},"algorithm-number-complement","整数的补数：位运算掩码解法","LeetCode 476 题，用掩码 XOR 实现整数补数，附 C++\u002FPython\u002FJava 三种实现及补数与补码的区别","2021-03-08",[11,421,13],"位运算",1374,[]]