[pull] main from doocs:main

solution/1000-1099/1039.Minimum Score Triangulation of Polygon/Solution.py

@@ -4,6 +4,9 @@ def minScoreTriangulation(self, values: List[int]) -> int:
         def dfs(i: int, j: int) -> int:
             if i + 1 == j:
                 return 0
-            return min(dfs(i, k) + dfs(k, j) + values[i] * values[k] * values[j] for k in range(i + 1, j))
+            return min(
+                dfs(i, k) + dfs(k, j) + values[i] * values[k] * values[j]
+                for k in range(i + 1, j)
+            )
 
         return dfs(0, len(values) - 1)

solution/1000-1099/1053.Previous Permutation With One Swap/README.md

@@ -142,6 +142,27 @@ func prevPermOpt1(arr []int) []int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function prevPermOpt1(arr: number[]): number[] {
+    const n = arr.length;
+    for (let i = n - 1; i > 0; --i) {
+        if (arr[i - 1] > arr[i]) {
+            for (let j = n - 1; j > i - 1; --j) {
+                if (arr[j] < arr[i - 1] && arr[j] !== arr[j - 1]) {
+                    const t = arr[i - 1];
+                    arr[i - 1] = arr[j];
+                    arr[j] = t;
+                    return arr;
+                }
+            }
+        }
+    }
+    return arr;
+}
+```
+
 ### **...**
 
 ```

solution/1000-1099/1053.Previous Permutation With One Swap/README_EN.md

@@ -124,6 +124,27 @@ func prevPermOpt1(arr []int) []int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function prevPermOpt1(arr: number[]): number[] {
+    const n = arr.length;
+    for (let i = n - 1; i > 0; --i) {
+        if (arr[i - 1] > arr[i]) {
+            for (let j = n - 1; j > i - 1; --j) {
+                if (arr[j] < arr[i - 1] && arr[j] !== arr[j - 1]) {
+                    const t = arr[i - 1];
+                    arr[i - 1] = arr[j];
+                    arr[j] = t;
+                    return arr;
+                }
+            }
+        }
+    }
+    return arr;
+}
+```
+
 ### **...**
 
 ```

solution/1000-1099/1053.Previous Permutation With One Swap/Solution.ts

@@ -0,0 +1,16 @@
+function prevPermOpt1(arr: number[]): number[] {
+    const n = arr.length;
+    for (let i = n - 1; i > 0; --i) {
+        if (arr[i - 1] > arr[i]) {
+            for (let j = n - 1; j > i - 1; --j) {
+                if (arr[j] < arr[i - 1] && arr[j] !== arr[j - 1]) {
+                    const t = arr[i - 1];
+                    arr[i - 1] = arr[j];
+                    arr[j] = t;
+                    return arr;
+                }
+            }
+        }
+    }
+    return arr;
+}

solution/1600-1699/1637.Widest Vertical Area Between Two Points Containing No Points/Solution.py

@@ -17,4 +17,4 @@ def maxWidthOfVerticalArea(self, points: List[List[int]]) -> int:
                 continue
             ans = max(ans, curmin - prev)
             prev = curmax
-        return ans
+        return ans

solution/1600-1699/1644.Lowest Common Ancestor of a Binary Tree II/Solution.py

@@ -5,8 +5,11 @@
 #         self.left = None
 #         self.right = None
 
+
 class Solution:
-    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
+    def lowestCommonAncestor(
+        self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode'
+    ) -> 'TreeNode':
         def dfs(root, p, q):
             if root is None:
                 return False

solution/1800-1899/1868.Product of Two Run-Length Encoded Arrays/Solution.py

@@ -1,5 +1,7 @@
 class Solution:
-    def findRLEArray(self, encoded1: List[List[int]], encoded2: List[List[int]]) -> List[List[int]]:
+    def findRLEArray(
+        self, encoded1: List[List[int]], encoded2: List[List[int]]
+    ) -> List[List[int]]:
         ans = []
         j = 0
         for vi, fi in encoded1:

solution/2400-2499/2428.Maximum Sum of an Hourglass/Solution.py

@@ -5,7 +5,8 @@ def maxSum(self, grid: List[List[int]]) -> int:
         for i in range(1, m - 1):
             for j in range(1, n - 1):
                 s = -grid[i][j - 1] - grid[i][j + 1]
-                s += sum(grid[x][y] for x in range(i - 1, i + 2)
-                         for y in range(j - 1, j + 2))
+                s += sum(
+                    grid[x][y] for x in range(i - 1, i + 2) for y in range(j - 1, j + 2)
+                )
                 ans = max(ans, s)
         return ans

solution/2500-2599/2550.Count Collisions of Monkeys on a Polygon/README.md

@@ -8,7 +8,7 @@
 
 <p>现在有一个正凸多边形，其上共有 <code>n</code> 个顶点。顶点按顺时针方向从 <code>0</code> 到 <code>n - 1</code> 依次编号。每个顶点上 <strong>正好有一只猴子</strong> 。下图中是一个 6 个顶点的凸多边形。</p>
 
-<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2500-2599/2550.Count%20Collisions%20of%20Monkeys%20on%20a%20Polygon/images/hexagon.jpg" style="width: 300px; height: 293px;"></p>
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2500-2599/2550.Count%20Collisions%20of%20Monkeys%20on%20a%20Polygon/images/hexagon.jpg" style="width: 300px; height: 293px;" /></p>
 
 <p>每个猴子同时移动到相邻的顶点。顶点 <code>i</code> 的相邻顶点可以是：</p>
 
@@ -17,7 +17,7 @@
 	<li>逆时针方向的顶点 <code>(i - 1 + n) % n</code> 。</li>
 </ul>
 
-<p>如果移动后至少有两个猴子位于同一顶点，则会发生 <strong>碰撞</strong> 。</p>
+<p>如果移动后至少有两只猴子停留在同一个顶点上或者相交在一条边上，则会发生 <strong>碰撞</strong> 。</p>
 
 <p>返回猴子至少发生 <strong>一次碰撞 </strong>的移动方法数。由于答案可能非常大，请返回对 <code>10<sup>9</sup>+7</code> 取余后的结果。</p>
 
@@ -27,7 +27,8 @@
 
 <p><strong>示例 1：</strong></p>
 
-<pre><strong>输入：</strong>n = 3
+<pre>
+<strong>输入：</strong>n = 3
 <strong>输出：</strong>6
 <strong>解释：</strong>共计 8 种移动方式。
 下面列出两种会发生碰撞的方式：
@@ -38,7 +39,8 @@
 
 <p><strong>示例 2：</strong></p>
 
-<pre><strong>输入：</strong>n = 4
+<pre>
+<strong>输入：</strong>n = 4
 <strong>输出：</strong>14
 <strong>解释：</strong>可以证明，有 14 种让猴子碰撞的方法。</pre>
 

solution/2600-2699/2600.K Items With the Maximum Sum/Solution.py

@@ -1,5 +1,7 @@
 class Solution:
-    def kItemsWithMaximumSum(self, numOnes: int, numZeros: int, numNegOnes: int, k: int) -> int:
+    def kItemsWithMaximumSum(
+        self, numOnes: int, numZeros: int, numNegOnes: int, k: int
+    ) -> int:
         if numOnes >= k:
             return k
         k -= numOnes