Merge branch 'main' of github.com:doocs/leetcode into feature/lc-1769

Author: rain84

images/starcharts.svg

@@ -271,6 +271,57 @@ function conveyorBelt(matrix: string[], start: number[], end: number[]): number
 }
 ```
 
+#### Swift
+
+```swift
+class Solution {
+    func conveyorBelt(_ matrix: [String], _ start: [Int], _ end: [Int]) -> Int {
+        let directions: [(Int, Int)] = [(-1, 0), (0, 1), (1, 0), (0, -1)]
+        let directionMap: [Character: Int] = ["^": 0, ">": 1, "v": 2, "<": 3]
+
+        let rows = matrix.count
+        let cols = matrix[0].count
+
+        var dist = Array(repeating: Array(repeating: Int.max, count: cols), count: rows)
+        var deque: [(Int, Int)] = []
+
+        dist[start[0]][start[1]] = 0
+        deque.append((start[0], start[1]))
+
+        while !deque.isEmpty {
+            let (i, j) = deque.removeFirst()
+
+            if i == end[0] && j == end[1] {
+                return dist[i][j]
+            }
+
+            for (k, (di, dj)) in directions.enumerated() {
+                let ni = i + di
+                let nj = j + dj
+
+                if ni >= 0 && ni < rows && nj >= 0 && nj < cols {
+                    let currentChar = matrix[i][matrix[i].index(matrix[i].startIndex, offsetBy: j)]
+                    let additionalCost = directionMap[currentChar] == k ? 0 : 1
+                    let newDist = dist[i][j] + additionalCost
+
+                    if newDist < dist[ni][nj] {
+                        dist[ni][nj] = newDist
+
+                        if additionalCost == 0 {
+                            deque.insert((ni, nj), at: 0)
+                        } else {
+                            deque.append((ni, nj))
+                        }
+                    }
+                }
+            }
+        }
+
+        return -1
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

lcp/LCP 56. 信物传送/README.md

@@ -0,0 +1,46 @@
+class Solution {
+    func conveyorBelt(_ matrix: [String], _ start: [Int], _ end: [Int]) -> Int {
+        let directions: [(Int, Int)] = [(-1, 0), (0, 1), (1, 0), (0, -1)]
+        let directionMap: [Character: Int] = ["^": 0, ">": 1, "v": 2, "<": 3]
+        
+        let rows = matrix.count
+        let cols = matrix[0].count
+        
+        var dist = Array(repeating: Array(repeating: Int.max, count: cols), count: rows)
+        var deque: [(Int, Int)] = []
+        
+        dist[start[0]][start[1]] = 0
+        deque.append((start[0], start[1]))
+        
+        while !deque.isEmpty {
+            let (i, j) = deque.removeFirst()
+            
+            if i == end[0] && j == end[1] {
+                return dist[i][j]
+            }
+            
+            for (k, (di, dj)) in directions.enumerated() {
+                let ni = i + di
+                let nj = j + dj
+                
+                if ni >= 0 && ni < rows && nj >= 0 && nj < cols {
+                    let currentChar = matrix[i][matrix[i].index(matrix[i].startIndex, offsetBy: j)]
+                    let additionalCost = directionMap[currentChar] == k ? 0 : 1
+                    let newDist = dist[i][j] + additionalCost
+                    
+                    if newDist < dist[ni][nj] {
+                        dist[ni][nj] = newDist
+                        
+                        if additionalCost == 0 {
+                            deque.insert((ni, nj), at: 0)
+                        } else {
+                            deque.append((ni, nj))
+                        }
+                    }
+                }
+            }
+        }
+        
+        return -1
+    }
+}

lcp/LCP 56. 信物传送/Solution.swift

@@ -183,6 +183,31 @@ impl Solution {
 }
 ```
 
+#### Swift
+
+```swift
+class Solution {
+    func temperatureTrend(_ temperatureA: [Int], _ temperatureB: [Int]) -> Int {
+        var maxTrend = 0
+        var currentTrend = 0
+
+        for i in 0..<temperatureA.count - 1 {
+            let changeA = temperatureA[i + 1] - temperatureA[i]
+            let changeB = temperatureB[i + 1] - temperatureB[i]
+
+            if (changeA == 0 && changeB == 0) || (changeA * changeB > 0) {
+                currentTrend += 1
+                maxTrend = max(maxTrend, currentTrend)
+            } else {
+                currentTrend = 0
+            }
+        }
+
+        return maxTrend
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

lcp/LCP 61. 气温变化趋势/README.md

@@ -0,0 +1,20 @@
+class Solution {
+    func temperatureTrend(_ temperatureA: [Int], _ temperatureB: [Int]) -> Int {
+        var maxTrend = 0
+        var currentTrend = 0
+        
+        for i in 0..<temperatureA.count - 1 {
+            let changeA = temperatureA[i + 1] - temperatureA[i]
+            let changeB = temperatureB[i + 1] - temperatureB[i]
+            
+            if (changeA == 0 && changeB == 0) || (changeA * changeB > 0) {
+                currentTrend += 1
+                maxTrend = max(maxTrend, currentTrend)
+            } else {
+                currentTrend = 0
+            }
+        }
+        
+        return maxTrend
+    }
+}

lcp/LCP 61. 气温变化趋势/Solution.swift

@@ -189,6 +189,32 @@ func crackSafe(n int, k int) string {
 }
 ```
 
+#### TypeScript
+
+```ts
+function crackSafe(n: number, k: number): string {
+    function dfs(u: number): void {
+        for (let x = 0; x < k; x++) {
+            const e = u * 10 + x;
+            if (!vis.has(e)) {
+                vis.add(e);
+                const v = e % mod;
+                dfs(v);
+                ans.push(x.toString());
+            }
+        }
+    }
+
+    const mod = Math.pow(10, n - 1);
+    const vis = new Set<number>();
+    const ans: string[] = [];
+
+    dfs(0);
+    ans.push('0'.repeat(n - 1));
+    return ans.join('');
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0700-0799/0753.Cracking the Safe/README.md

@@ -76,7 +76,13 @@ Thus "01100" will unlock the safe. "10011", and "11001&
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Eulerian Circuit
+
+We can construct a directed graph based on the description in the problem: each point is considered as a length $n-1$ $k$-string, and each edge carries a character from $0$ to $k-1$. If there is a directed edge $e$ from point $u$ to point $v$, and the character carried by $e$ is $c$, then the last $k-1$ characters of $u+c$ form the string $v$. At this point, the edge $u+c$ represents a password of length $n$.
+
+In this directed graph, there are $k^{n-1}$ points, each point has $k$ outgoing edges and $k$ incoming edges. Therefore, this directed graph has an Eulerian circuit, and the path traversed by the Eulerian circuit is the answer to the problem.
+
+The time complexity is $O(k^n)$, and the space complexity is $O(k^n)$.
 
 <!-- tabs:start -->
 
@@ -181,6 +187,32 @@ func crackSafe(n int, k int) string {
 }
 ```
 
+#### TypeScript
+
+```ts
+function crackSafe(n: number, k: number): string {
+    function dfs(u: number): void {
+        for (let x = 0; x < k; x++) {
+            const e = u * 10 + x;
+            if (!vis.has(e)) {
+                vis.add(e);
+                const v = e % mod;
+                dfs(v);
+                ans.push(x.toString());
+            }
+        }
+    }
+
+    const mod = Math.pow(10, n - 1);
+    const vis = new Set<number>();
+    const ans: string[] = [];
+
+    dfs(0);
+    ans.push('0'.repeat(n - 1));
+    return ans.join('');
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/0700-0799/0753.Cracking the Safe/README_EN.md

@@ -0,0 +1,21 @@
+function crackSafe(n: number, k: number): string {
+    function dfs(u: number): void {
+        for (let x = 0; x < k; x++) {
+            const e = u * 10 + x;
+            if (!vis.has(e)) {
+                vis.add(e);
+                const v = e % mod;
+                dfs(v);
+                ans.push(x.toString());
+            }
+        }
+    }
+
+    const mod = Math.pow(10, n - 1);
+    const vis = new Set<number>();
+    const ans: string[] = [];
+
+    dfs(0);
+    ans.push('0'.repeat(n - 1));
+    return ans.join('');
+}

solution/0700-0799/0753.Cracking the Safe/Solution.ts

@@ -68,15 +68,15 @@ tags:
 
 ### 方法一：数学分析
 
-由于对称性，每次可以选择向左或向右移动，因此，我们可以将 $target$ 统一取绝对值。
+由于对称性，每次可以选择向左或向右移动，因此，我们可以将 $\textit{target}$ 统一取绝对值。
 
 定义 $s$ 表示当前所处的位置，用变量 $k$ 记录移动的次数。初始时 $s$ 和 $k$ 均为 $0$。
 
-我们将 $s$ 一直循环累加，直到满足 $s\ge target$ 并且 $(s-target)\mod 2 = 0$，此时的移动次数 $k$ 就是答案，直接返回。
+我们将 $s$ 一直循环累加，直到满足 $s\ge \textit{target}$ 并且 $(s-\textit{target}) \bmod 2 = 0$，此时的移动次数 $k$ 就是答案，直接返回。
 
-为什么？因为如果 $s\ge target$ 且 $(s-target)\mod 2 = 0$，我们只需要把前面 $\frac{s-target}{2}$ 这个正整数变为负数，就能使得 $s$ 与 $target$ 相等。正整数变负数的过程，实际上是将移动的方向改变，但实际移动次数仍然不变。
+为什么？因为如果 $s\ge \textit{target}$ 且 $(s-\textit{target})\mod 2 = 0$，我们只需要把前面 $\frac{s-\textit{target}}{2}$ 这个正整数变为负数，就能使得 $s$ 与 $\textit{target}$ 相等。正整数变负数的过程，实际上是将移动的方向改变，但实际移动次数仍然不变。
 
-时间复杂度 $O(\sqrt{\left | target \right | })$，空间复杂度 $O(1)$。
+时间复杂度 $O(\sqrt{\left | \textit{target} \right | })$，空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 

solution/0700-0799/0754.Reach a Number/README.md

@@ -64,7 +64,17 @@ On the 2<sup>nd</sup> move, we step from 1 to 3 (2 steps).
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Mathematical Analysis
+
+Due to symmetry, each time we can choose to move left or right, so we can take the absolute value of $\textit{target}$.
+
+Define $s$ as the current position, and use the variable $k$ to record the number of moves. Initially, both $s$ and $k$ are $0$.
+
+We keep adding to $s$ in a loop until $s \ge \textit{target}$ and $(s - \textit{target}) \bmod 2 = 0$. At this point, the number of moves $k$ is the answer, and we return it directly.
+
+Why? Because if $s \ge \textit{target}$ and $(s - \textit{target}) \bmod 2 = 0$, we only need to change the sign of the positive integer $\frac{s - \textit{target}}{2}$ to negative, so that $s$ equals $\textit{target}$. Changing the sign of a positive integer essentially means changing the direction of the move, but the actual number of moves remains the same.
+
+The time complexity is $O(\sqrt{\left | \textit{target} \right | })$, and the space complexity is $O(1)$.
 
 <!-- tabs:start -->