Merge branch 'main' into main

Author: AE-Hertz

images/starcharts.svg

@@ -152,36 +152,30 @@ class Solution {
 #### C++
 
 ```cpp
-int f[100010][2][3];
-const int mod = 1e9 + 7;
-
 class Solution {
 public:
     int checkRecord(int n) {
-        this->n = n;
+        int f[n][2][3];
         memset(f, -1, sizeof(f));
-        return dfs(0, 0, 0);
-    }
-
-    int dfs(int i, int j, int k) {
-        if (i >= n) {
-            return 1;
-        }
-        if (f[i][j][k] != -1) {
-            return f[i][j][k];
-        }
-        int ans = dfs(i + 1, j, 0);
-        if (j == 0) {
-            ans = (ans + dfs(i + 1, j + 1, 0)) % mod;
-        }
-        if (k < 2) {
-            ans = (ans + dfs(i + 1, j, k + 1)) % mod;
-        }
-        return f[i][j][k] = ans;
+        const int mod = 1e9 + 7;
+        auto dfs = [&](auto&& dfs, int i, int j, int k) -> int {
+            if (i >= n) {
+                return 1;
+            }
+            if (f[i][j][k] != -1) {
+                return f[i][j][k];
+            }
+            int ans = dfs(dfs, i + 1, j, 0);
+            if (j == 0) {
+                ans = (ans + dfs(dfs, i + 1, j + 1, 0)) % mod;
+            }
+            if (k < 2) {
+                ans = (ans + dfs(dfs, i + 1, j, k + 1)) % mod;
+            }
+            return f[i][j][k] = ans;
+        };
+        return dfs(dfs, 0, 0, 0);
     }
-
-private:
-    int n;
 };
 ```
 

solution/0500-0599/0552.Student Attendance Record II/README.md

@@ -134,36 +134,30 @@ class Solution {
 #### C++
 
 ```cpp
-int f[100010][2][3];
-const int mod = 1e9 + 7;
-
 class Solution {
 public:
     int checkRecord(int n) {
-        this->n = n;
+        int f[n][2][3];
         memset(f, -1, sizeof(f));
-        return dfs(0, 0, 0);
-    }
-
-    int dfs(int i, int j, int k) {
-        if (i >= n) {
-            return 1;
-        }
-        if (f[i][j][k] != -1) {
-            return f[i][j][k];
-        }
-        int ans = dfs(i + 1, j, 0);
-        if (j == 0) {
-            ans = (ans + dfs(i + 1, j + 1, 0)) % mod;
-        }
-        if (k < 2) {
-            ans = (ans + dfs(i + 1, j, k + 1)) % mod;
-        }
-        return f[i][j][k] = ans;
+        const int mod = 1e9 + 7;
+        auto dfs = [&](auto&& dfs, int i, int j, int k) -> int {
+            if (i >= n) {
+                return 1;
+            }
+            if (f[i][j][k] != -1) {
+                return f[i][j][k];
+            }
+            int ans = dfs(dfs, i + 1, j, 0);
+            if (j == 0) {
+                ans = (ans + dfs(dfs, i + 1, j + 1, 0)) % mod;
+            }
+            if (k < 2) {
+                ans = (ans + dfs(dfs, i + 1, j, k + 1)) % mod;
+            }
+            return f[i][j][k] = ans;
+        };
+        return dfs(dfs, 0, 0, 0);
     }
-
-private:
-    int n;
 };
 ```
 

solution/0500-0599/0552.Student Attendance Record II/README_EN.md

@@ -1,31 +1,25 @@
-int f[100010][2][3];
-const int mod = 1e9 + 7;
-
 class Solution {
 public:
     int checkRecord(int n) {
-        this->n = n;
+        int f[n][2][3];
         memset(f, -1, sizeof(f));
-        return dfs(0, 0, 0);
+        const int mod = 1e9 + 7;
+        auto dfs = [&](auto&& dfs, int i, int j, int k) -> int {
+            if (i >= n) {
+                return 1;
+            }
+            if (f[i][j][k] != -1) {
+                return f[i][j][k];
+            }
+            int ans = dfs(dfs, i + 1, j, 0);
+            if (j == 0) {
+                ans = (ans + dfs(dfs, i + 1, j + 1, 0)) % mod;
+            }
+            if (k < 2) {
+                ans = (ans + dfs(dfs, i + 1, j, k + 1)) % mod;
+            }
+            return f[i][j][k] = ans;
+        };
+        return dfs(dfs, 0, 0, 0);
     }
-
-    int dfs(int i, int j, int k) {
-        if (i >= n) {
-            return 1;
-        }
-        if (f[i][j][k] != -1) {
-            return f[i][j][k];
-        }
-        int ans = dfs(i + 1, j, 0);
-        if (j == 0) {
-            ans = (ans + dfs(i + 1, j + 1, 0)) % mod;
-        }
-        if (k < 2) {
-            ans = (ans + dfs(i + 1, j, k + 1)) % mod;
-        }
-        return f[i][j][k] = ans;
-    }
-
-private:
-    int n;
-};
+};

solution/0500-0599/0552.Student Attendance Record II/Solution.cpp

@@ -9,36 +9,48 @@ tags:
 
 <!-- problem:start -->
 
-# [624. 数组列表中的最大距离 🔒](https://leetcode.cn/problems/maximum-distance-in-arrays)
+# [624. 数组列表中的最大距离](https://leetcode.cn/problems/maximum-distance-in-arrays)
 
 [English Version](/solution/0600-0699/0624.Maximum%20Distance%20in%20Arrays/README_EN.md)
 
 ## 题目描述
 
 <!-- description:start -->
 
-<p>给定&nbsp;<code>m</code>&nbsp;个数组，每个数组都已经按照升序排好序了。现在你需要从两个不同的数组中选择两个整数（每个数组选一个）并且计算它们的距离。两个整数&nbsp;<code>a</code>&nbsp;和&nbsp;<code>b</code>&nbsp;之间的距离定义为它们差的绝对值&nbsp;<code>|a-b|</code>&nbsp;。你的任务就是去找到最大距离</p>
+<p>给定&nbsp;<code>m</code>&nbsp;个数组，每个数组都已经按照升序排好序了。</p>
+
+<p>现在你需要从两个不同的数组中选择两个整数（每个数组选一个）并且计算它们的距离。两个整数&nbsp;<code>a</code>&nbsp;和&nbsp;<code>b</code>&nbsp;之间的距离定义为它们差的绝对值&nbsp;<code>|a-b|</code>&nbsp;。</p>
+
+<p>返回最大距离。</p>
 
 <p><strong>示例 1：</strong></p>
 
-<pre><strong>输入：</strong> 
-[[1,2,3],
- [4,5],
- [1,2,3]]
-<strong>输出：</strong> 4
+<pre>
+<strong>输入：</strong>[[1,2,3],[4,5],[1,2,3]]
+<strong>输出：</strong>4
 <strong>解释：</strong>
 一种得到答案 4 的方法是从第一个数组或者第三个数组中选择 1，同时从第二个数组中选择 5 。
 </pre>
 
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>arrays = [[1],[1]]
+<b>输出：</b>0
+</pre>
+
 <p>&nbsp;</p>
 
-<p><strong>注意：</strong></p>
+<p><strong>提示：</strong></p>
 
-<ol>
-	<li>每个给定数组至少会有 1 个数字。列表中至少有两个非空数组。</li>
-	<li><strong>所有</strong>&nbsp;<code>m</code>&nbsp;个数组中的数字总数目在范围 [2, 10000] 内。</li>
-	<li><code>m</code>&nbsp;个数组中所有整数的范围在 [-10000, 10000] 内。</li>
-</ol>
+<ul>
+	<li><code>m == arrays.length</code></li>
+	<li><code>2 &lt;= m &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= arrays[i].length &lt;= 500</code></li>
+	<li><code>-10<sup>4</sup> &lt;= arrays[i][j] &lt;= 10<sup>4</sup></code></li>
+	<li><code>arrays[i]</code>&nbsp;以&nbsp;<strong>升序</strong>&nbsp;排序。</li>
+	<li>所有数组中最多有&nbsp;<code>10<sup>5</sup></code> 个整数。</li>
+</ul>
 
 <p>&nbsp;</p>
 
@@ -136,6 +148,90 @@ func abs(x int) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function maxDistance(arrays: number[][]): number {
+    const n = arrays.length;
+    let res = 0;
+    let [min, max] = [Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
+
+    for (let i = 0; i < n; i++) {
+        const a = arrays[i];
+        res = Math.max(Math.max(a.at(-1)! - min, max - a[0]), res);
+        min = Math.min(min, a[0]);
+        max = Math.max(max, a.at(-1)!);
+    }
+
+    return res;
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {number[][]} arrays
+ * @return {number}
+ */
+var maxDistance = function (arrays) {
+    const n = arrays.length;
+    let res = 0;
+    let [min, max] = [Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
+
+    for (let i = 0; i < n; i++) {
+        const a = arrays[i];
+        res = Math.max(Math.max(a.at(-1) - min, max - a[0]), res);
+        min = Math.min(min, a[0]);
+        max = Math.max(max, a.at(-1));
+    }
+
+    return res;
+};
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- solution:start -->
+
+### 方法二：一行
+
+<!-- tabs:start -->
+
+#### TypeScript
+
+```ts
+const maxDistance = (arrays: number[][]): number =>
+    arrays.reduce(
+        ([res, min, max], a) => [
+            Math.max(Math.max(a.at(-1)! - min, max - a[0]), res),
+            Math.min(min, a[0]),
+            Math.max(max, a.at(-1)!),
+        ],
+        [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY],
+    )[0];
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {number[][]} arrays
+ * @return {number}
+ */
+var maxDistance = arrays =>
+    arrays.reduce(
+        ([res, min, max], a) => [
+            Math.max(Math.max(a.at(-1) - min, max - a[0]), res),
+            Math.min(min, a[0]),
+            Math.max(max, a.at(-1)),
+        ],
+        [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY],
+    )[0];
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->