feat: update lc problems (#3089)

Author: yanglbme

solution/0100-0199/0130.Surrounded Regions/README_EN.md

@@ -20,29 +20,36 @@ tags:
 
 <!-- description:start -->
 
-<p>Given an <code>m x n</code> matrix <code>board</code> containing <code>&#39;X&#39;</code> and <code>&#39;O&#39;</code>, <em>capture all regions that are 4-directionally&nbsp;surrounded by</em> <code>&#39;X&#39;</code>.</p>
+<p>You are given an <code>m x n</code> matrix <code>board</code> containing <strong>letters</strong> <code>&#39;X&#39;</code> and <code>&#39;O&#39;</code>, <strong>capture regions</strong> that are <strong>surrounded</strong>:</p>
 
-<p>A region is <strong>captured</strong> by flipping all <code>&#39;O&#39;</code>s into <code>&#39;X&#39;</code>s in that surrounded region.</p>
+<ul>
+	<li><strong>Connect</strong>: A cell is connected to adjacent cells horizontally or vertically.</li>
+	<li><strong>Region</strong>: To form a region <strong>connect every</strong> <code>&#39;O&#39;</code> cell.</li>
+	<li><strong>Surround</strong>: The region is surrounded with <code>&#39;X&#39;</code> cells if you can <strong>connect the region </strong>with <code>&#39;X&#39;</code> cells and none of the region cells are on the edge of the <code>board</code>.</li>
+</ul>
+
+<p>A <strong>surrounded region is captured</strong> by replacing all <code>&#39;O&#39;</code>s with <code>&#39;X&#39;</code>s in the input matrix <code>board</code>.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
-<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0130.Surrounded%20Regions/images/xogrid.jpg" style="width: 550px; height: 237px;" />
-<pre>
-<strong>Input:</strong> board = [[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;X&quot;,&quot;X&quot;]]
-<strong>Output:</strong> [[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;X&quot;,&quot;X&quot;]]
-<strong>Explanation:</strong> Notice that an &#39;O&#39; should not be flipped if:
-- It is on the border, or
-- It is adjacent to an &#39;O&#39; that should not be flipped.
-The bottom &#39;O&#39; is on the border, so it is not flipped.
-The other three &#39;O&#39; form a surrounded region, so they are flipped.
-</pre>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">board = [[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;X&quot;,&quot;X&quot;]]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;X&quot;,&quot;X&quot;]]</span></p>
+
+<p><strong>Explanation:</strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0130.Surrounded%20Regions/images/xogrid.jpg" style="width: 367px; height: 158px;" />
+<p>In the above diagram, the bottom region is not captured because it is on the edge of the board and cannot be surrounded.</p>
+</div>
 
 <p><strong class="example">Example 2:</strong></p>
 
-<pre>
-<strong>Input:</strong> board = [[&quot;X&quot;]]
-<strong>Output:</strong> [[&quot;X&quot;]]
-</pre>
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">board = [[&quot;X&quot;]]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[[&quot;X&quot;]]</span></p>
+</div>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>

solution/0400-0499/0419.Battleships in a Board/README.md

@@ -59,7 +59,13 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：直接遍历
+
+我们可以遍历矩阵，找到每个战舰的左上角，即当前位置为 `X` 且上方和左方都不是 `X` 的位置，将答案加一。
+
+遍历结束后，返回答案即可。
+
+时间复杂度 $O(m \times n)$，其中 $m$ 和 $n$ 分别是矩阵的行数和列数。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -118,9 +124,15 @@ public:
         int ans = 0;
         for (int i = 0; i < m; ++i) {
             for (int j = 0; j < n; ++j) {
-                if (board[i][j] == '.') continue;
-                if (i > 0 && board[i - 1][j] == 'X') continue;
-                if (j > 0 && board[i][j - 1] == 'X') continue;
+                if (board[i][j] == '.') {
+                    continue;
+                }
+                if (i > 0 && board[i - 1][j] == 'X') {
+                    continue;
+                }
+                if (j > 0 && board[i][j - 1] == 'X') {
+                    continue;
+                }
                 ++ans;
             }
         }
@@ -132,12 +144,10 @@ public:
 #### Go
 
 ```go
-func countBattleships(board [][]byte) int {
-	m, n := len(board), len(board[0])
-	ans := 0
-	for i := 0; i < m; i++ {
-		for j := 0; j < n; j++ {
-			if board[i][j] == '.' {
+func countBattleships(board [][]byte) (ans int) {
+	for i, row := range board {
+		for j, c := range row {
+			if c == '.' {
 				continue
 			}
 			if i > 0 && board[i-1][j] == 'X' {
@@ -149,7 +159,32 @@ func countBattleships(board [][]byte) int {
 			ans++
 		}
 	}
-	return ans
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function countBattleships(board: string[][]): number {
+    const m = board.length;
+    const n = board[0].length;
+    let ans = 0;
+    for (let i = 0; i < m; ++i) {
+        for (let j = 0; j < n; ++j) {
+            if (board[i][j] === '.') {
+                continue;
+            }
+            if (i && board[i - 1][j] === 'X') {
+                continue;
+            }
+            if (j && board[i][j - 1] === 'X') {
+                continue;
+            }
+            ++ans;
+        }
+    }
+    return ans;
 }
 ```
 

solution/0400-0499/0419.Battleships in a Board/README_EN.md

@@ -56,7 +56,13 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Direct Iteration
+
+We can iterate through the matrix, find the top-left corner of each battleship, i.e., the position where the current position is `X` and both the top and left are not `X`, and increment the answer by one.
+
+After the iteration ends, return the answer.
+
+The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the matrix, respectively. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -115,9 +121,15 @@ public:
         int ans = 0;
         for (int i = 0; i < m; ++i) {
             for (int j = 0; j < n; ++j) {
-                if (board[i][j] == '.') continue;
-                if (i > 0 && board[i - 1][j] == 'X') continue;
-                if (j > 0 && board[i][j - 1] == 'X') continue;
+                if (board[i][j] == '.') {
+                    continue;
+                }
+                if (i > 0 && board[i - 1][j] == 'X') {
+                    continue;
+                }
+                if (j > 0 && board[i][j - 1] == 'X') {
+                    continue;
+                }
                 ++ans;
             }
         }
@@ -129,12 +141,10 @@ public:
 #### Go
 
 ```go
-func countBattleships(board [][]byte) int {
-	m, n := len(board), len(board[0])
-	ans := 0
-	for i := 0; i < m; i++ {
-		for j := 0; j < n; j++ {
-			if board[i][j] == '.' {
+func countBattleships(board [][]byte) (ans int) {
+	for i, row := range board {
+		for j, c := range row {
+			if c == '.' {
 				continue
 			}
 			if i > 0 && board[i-1][j] == 'X' {
@@ -146,7 +156,32 @@ func countBattleships(board [][]byte) int {
 			ans++
 		}
 	}
-	return ans
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function countBattleships(board: string[][]): number {
+    const m = board.length;
+    const n = board[0].length;
+    let ans = 0;
+    for (let i = 0; i < m; ++i) {
+        for (let j = 0; j < n; ++j) {
+            if (board[i][j] === '.') {
+                continue;
+            }
+            if (i && board[i - 1][j] === 'X') {
+                continue;
+            }
+            if (j && board[i][j - 1] === 'X') {
+                continue;
+            }
+            ++ans;
+        }
+    }
+    return ans;
 }
 ```
 

solution/0400-0499/0419.Battleships in a Board/Solution.cpp

@@ -5,9 +5,15 @@ class Solution {
         int ans = 0;
         for (int i = 0; i < m; ++i) {
             for (int j = 0; j < n; ++j) {
-                if (board[i][j] == '.') continue;
-                if (i > 0 && board[i - 1][j] == 'X') continue;
-                if (j > 0 && board[i][j - 1] == 'X') continue;
+                if (board[i][j] == '.') {
+                    continue;
+                }
+                if (i > 0 && board[i - 1][j] == 'X') {
+                    continue;
+                }
+                if (j > 0 && board[i][j - 1] == 'X') {
+                    continue;
+                }
                 ++ans;
             }
         }

solution/0400-0499/0419.Battleships in a Board/Solution.go

@@ -1,9 +1,7 @@
-func countBattleships(board [][]byte) int {
-	m, n := len(board), len(board[0])
-	ans := 0
-	for i := 0; i < m; i++ {
-		for j := 0; j < n; j++ {
-			if board[i][j] == '.' {
+func countBattleships(board [][]byte) (ans int) {
+	for i, row := range board {
+		for j, c := range row {
+			if c == '.' {
 				continue
 			}
 			if i > 0 && board[i-1][j] == 'X' {
@@ -15,5 +13,5 @@ func countBattleships(board [][]byte) int {
 			ans++
 		}
 	}
-	return ans
+	return
 }

solution/0400-0499/0419.Battleships in a Board/Solution.ts

@@ -0,0 +1,20 @@
+function countBattleships(board: string[][]): number {
+    const m = board.length;
+    const n = board[0].length;
+    let ans = 0;
+    for (let i = 0; i < m; ++i) {
+        for (let j = 0; j < n; ++j) {
+            if (board[i][j] === '.') {
+                continue;
+            }
+            if (i && board[i - 1][j] === 'X') {
+                continue;
+            }
+            if (j && board[i][j - 1] === 'X') {
+                continue;
+            }
+            ++ans;
+        }
+    }
+    return ans;
+}

solution/0800-0899/0853.Car Fleet/README_EN.md

@@ -19,48 +19,59 @@ tags:
 
 <!-- description:start -->
 
-<p>There are <code>n</code> cars going to the same destination along a one-lane road. The destination is <code>target</code> miles away.</p>
+<p>There are <code>n</code> cars at given miles away from the starting mile 0, traveling to reach the mile <code>target</code>.</p>
 
-<p>You are given two integer array <code>position</code> and <code>speed</code>, both of length <code>n</code>, where <code>position[i]</code> is the position of the <code>i<sup>th</sup></code> car and <code>speed[i]</code> is the speed of the <code>i<sup>th</sup></code> car (in miles per hour).</p>
+<p>You are given two integer array <code>position</code> and <code>speed</code>, both of length <code>n</code>, where <code>position[i]</code> is the starting mile of the <code>i<sup>th</sup></code> car and <code>speed[i]</code> is the speed of the <code>i<sup>th</sup></code> car in miles per hour.</p>
 
-<p>A car can never pass another car ahead of it, but it can catch up to it&nbsp;and drive bumper to bumper <strong>at the same speed</strong>. The faster car will <strong>slow down</strong> to match the slower car&#39;s speed. The distance between these two cars is ignored (i.e., they are assumed to have the same position).</p>
+<p>A car cannot pass another car, but it can catch up and then travel next to it at the speed of the slower car.</p>
 
-<p>A <strong>car fleet</strong> is some non-empty set of cars driving at the same position and same speed. Note that a single car is also a car fleet.</p>
+<p>A <strong>car fleet</strong> is a car or cars driving next to each other. The speed of the car fleet is the <strong>minimum</strong> speed of any car in the fleet.</p>
 
-<p>If a car catches up to a car fleet right at the destination point, it will still be considered as one car fleet.</p>
+<p>If a car catches up to a car fleet at the mile <code>target</code>, it will still be considered as part of the car fleet.</p>
 
-<p>Return <em>the <strong>number of car fleets</strong> that will arrive at the destination</em>.</p>
+<p>Return the number of car fleets that will arrive at the destination.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 
-<pre>
-<strong>Input:</strong> target = 12, position = [10,8,0,5,3], speed = [2,4,1,1,3]
-<strong>Output:</strong> 3
-<strong>Explanation:</strong>
-The cars starting at 10 (speed 2) and 8 (speed 4) become a fleet, meeting each other at 12.
-The car starting at 0 does not catch up to any other car, so it is a fleet by itself.
-The cars starting at 5 (speed 1) and 3 (speed 3) become a fleet, meeting each other at 6. The fleet moves at speed 1 until it reaches target.
-Note that no other cars meet these fleets before the destination, so the answer is 3.
-</pre>
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">target = 12, position = [10,8,0,5,3], speed = [2,4,1,1,3]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">3</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>The cars starting at 10 (speed 2) and 8 (speed 4) become a fleet, meeting each other at 12. The fleet forms at <code>target</code>.</li>
+	<li>The car starting at 0 (speed 1) does not catch up to any other car, so it is a fleet by itself.</li>
+	<li>The cars starting at 5 (speed 1) and 3 (speed 3) become a fleet, meeting each other at 6. The fleet moves at speed 1 until it reaches <code>target</code>.</li>
+</ul>
+</div>
 
 <p><strong class="example">Example 2:</strong></p>
 
-<pre>
-<strong>Input:</strong> target = 10, position = [3], speed = [3]
-<strong>Output:</strong> 1
-<strong>Explanation:</strong> There is only one car, hence there is only one fleet.
-</pre>
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">target = 10, position = [3], speed = [3]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+
+<p><strong>Explanation:</strong></p>
+There is only one car, hence there is only one fleet.</div>
 
 <p><strong class="example">Example 3:</strong></p>
 
-<pre>
-<strong>Input:</strong> target = 100, position = [0,2,4], speed = [4,2,1]
-<strong>Output:</strong> 1
-<strong>Explanation:</strong>
-The cars starting at 0 (speed 4) and 2 (speed 2) become a fleet, meeting each other at 4. The fleet moves at speed 2.
-Then, the fleet (speed 2) and the car starting at 4 (speed 1) become one fleet, meeting each other at 6. The fleet moves at speed 1 until it reaches target.
-</pre>
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">target = 100, position = [0,2,4], speed = [4,2,1]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>The cars starting at 0 (speed 4) and 2 (speed 2) become a fleet, meeting each other at 4. The car starting at 4 (speed 1) travels to 5.</li>
+	<li>Then, the fleet at 4 (speed 2) and the car at position 5 (speed 1) become one fleet, meeting each other at 6. The fleet moves at speed 1 until it reaches <code>target</code>.</li>
+</ul>
+</div>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>