feat: add solutions to lc problems: No.0541,0542

solution/0500-0599/0541.Reverse String II/README.md

@@ -56,7 +56,11 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：双指针
+
+我们可以遍历字符串 $\textit{s}$，每次遍历 $\textit{2k}$ 个字符，然后利用双指针技巧，对这 $\textit{2k}$ 个字符中的前 $\textit{k}$ 个字符进行反转。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串 $\textit{s}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -65,26 +69,27 @@ tags:
 ```python
 class Solution:
     def reverseStr(self, s: str, k: int) -> str:
-        t = list(s)
-        for i in range(0, len(t), k << 1):
-            t[i : i + k] = reversed(t[i : i + k])
-        return ''.join(t)
+        cs = list(s)
+        for i in range(0, len(cs), 2 * k):
+            cs[i : i + k] = reversed(cs[i : i + k])
+        return "".join(cs)
 ```
 
 #### Java
 
 ```java
 class Solution {
     public String reverseStr(String s, int k) {
-        char[] chars = s.toCharArray();
-        for (int i = 0; i < chars.length; i += (k << 1)) {
-            for (int st = i, ed = Math.min(chars.length - 1, i + k - 1); st < ed; ++st, --ed) {
-                char t = chars[st];
-                chars[st] = chars[ed];
-                chars[ed] = t;
+        char[] cs = s.toCharArray();
+        int n = cs.length;
+        for (int i = 0; i < n; i += k * 2) {
+            for (int l = i, r = Math.min(i + k - 1, n - 1); l < r; ++l, --r) {
+                char t = cs[l];
+                cs[l] = cs[r];
+                cs[r] = t;
             }
         }
-        return new String(chars);
+        return new String(cs);
     }
 }
 ```
@@ -95,7 +100,8 @@ class Solution {
 class Solution {
 public:
     string reverseStr(string s, int k) {
-        for (int i = 0, n = s.size(); i < n; i += (k << 1)) {
+        int n = s.size();
+        for (int i = 0; i < n; i += 2 * k) {
             reverse(s.begin() + i, s.begin() + min(i + k, n));
         }
         return s;
@@ -107,13 +113,29 @@ public:
 
 ```go
 func reverseStr(s string, k int) string {
-	t := []byte(s)
-	for i := 0; i < len(t); i += (k << 1) {
-		for st, ed := i, min(i+k-1, len(t)-1); st < ed; st, ed = st+1, ed-1 {
-			t[st], t[ed] = t[ed], t[st]
+	cs := []byte(s)
+	n := len(cs)
+	for i := 0; i < n; i += 2 * k {
+		for l, r := i, min(i+k-1, n-1); l < r; l, r = l+1, r-1 {
+			cs[l], cs[r] = cs[r], cs[l]
 		}
 	}
-	return string(t)
+	return string(cs)
+}
+```
+
+#### TypeScript
+
+```ts
+function reverseStr(s: string, k: number): string {
+    const n = s.length;
+    const cs = s.split('');
+    for (let i = 0; i < n; i += 2 * k) {
+        for (let l = i, r = Math.min(i + k - 1, n - 1); l < r; l++, r--) {
+            [cs[l], cs[r]] = [cs[r], cs[l]];
+        }
+    }
+    return cs.join('');
 }
 ```
 

solution/0500-0599/0541.Reverse String II/README_EN.md

@@ -44,7 +44,11 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Two Pointers
+
+We can traverse the string $\textit{s}$, iterating over every $\textit{2k}$ characters, and then use the two-pointer technique to reverse the first $\textit{k}$ characters among these $\textit{2k}$ characters.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $\textit{s}$.
 
 <!-- tabs:start -->
 
@@ -53,26 +57,27 @@ tags:
 ```python
 class Solution:
     def reverseStr(self, s: str, k: int) -> str:
-        t = list(s)
-        for i in range(0, len(t), k << 1):
-            t[i : i + k] = reversed(t[i : i + k])
-        return ''.join(t)
+        cs = list(s)
+        for i in range(0, len(cs), 2 * k):
+            cs[i : i + k] = reversed(cs[i : i + k])
+        return "".join(cs)
 ```
 
 #### Java
 
 ```java
 class Solution {
     public String reverseStr(String s, int k) {
-        char[] chars = s.toCharArray();
-        for (int i = 0; i < chars.length; i += (k << 1)) {
-            for (int st = i, ed = Math.min(chars.length - 1, i + k - 1); st < ed; ++st, --ed) {
-                char t = chars[st];
-                chars[st] = chars[ed];
-                chars[ed] = t;
+        char[] cs = s.toCharArray();
+        int n = cs.length;
+        for (int i = 0; i < n; i += k * 2) {
+            for (int l = i, r = Math.min(i + k - 1, n - 1); l < r; ++l, --r) {
+                char t = cs[l];
+                cs[l] = cs[r];
+                cs[r] = t;
             }
         }
-        return new String(chars);
+        return new String(cs);
     }
 }
 ```
@@ -83,7 +88,8 @@ class Solution {
 class Solution {
 public:
     string reverseStr(string s, int k) {
-        for (int i = 0, n = s.size(); i < n; i += (k << 1)) {
+        int n = s.size();
+        for (int i = 0; i < n; i += 2 * k) {
             reverse(s.begin() + i, s.begin() + min(i + k, n));
         }
         return s;
@@ -95,13 +101,29 @@ public:
 
 ```go
 func reverseStr(s string, k int) string {
-	t := []byte(s)
-	for i := 0; i < len(t); i += (k << 1) {
-		for st, ed := i, min(i+k-1, len(t)-1); st < ed; st, ed = st+1, ed-1 {
-			t[st], t[ed] = t[ed], t[st]
+	cs := []byte(s)
+	n := len(cs)
+	for i := 0; i < n; i += 2 * k {
+		for l, r := i, min(i+k-1, n-1); l < r; l, r = l+1, r-1 {
+			cs[l], cs[r] = cs[r], cs[l]
 		}
 	}
-	return string(t)
+	return string(cs)
+}
+```
+
+#### TypeScript
+
+```ts
+function reverseStr(s: string, k: number): string {
+    const n = s.length;
+    const cs = s.split('');
+    for (let i = 0; i < n; i += 2 * k) {
+        for (let l = i, r = Math.min(i + k - 1, n - 1); l < r; l++, r--) {
+            [cs[l], cs[r]] = [cs[r], cs[l]];
+        }
+    }
+    return cs.join('');
 }
 ```
 

solution/0500-0599/0541.Reverse String II/Solution.cpp

@@ -1,7 +1,8 @@
 class Solution {
 public:
     string reverseStr(string s, int k) {
-        for (int i = 0, n = s.size(); i < n; i += (k << 1)) {
+        int n = s.size();
+        for (int i = 0; i < n; i += 2 * k) {
             reverse(s.begin() + i, s.begin() + min(i + k, n));
         }
         return s;

solution/0500-0599/0541.Reverse String II/Solution.go

@@ -1,9 +1,10 @@
 func reverseStr(s string, k int) string {
-	t := []byte(s)
-	for i := 0; i < len(t); i += (k << 1) {
-		for st, ed := i, min(i+k-1, len(t)-1); st < ed; st, ed = st+1, ed-1 {
-			t[st], t[ed] = t[ed], t[st]
+	cs := []byte(s)
+	n := len(cs)
+	for i := 0; i < n; i += 2 * k {
+		for l, r := i, min(i+k-1, n-1); l < r; l, r = l+1, r-1 {
+			cs[l], cs[r] = cs[r], cs[l]
 		}
 	}
-	return string(t)
+	return string(cs)
 }

solution/0500-0599/0541.Reverse String II/Solution.java

@@ -1,13 +1,14 @@
 class Solution {
     public String reverseStr(String s, int k) {
-        char[] chars = s.toCharArray();
-        for (int i = 0; i < chars.length; i += (k << 1)) {
-            for (int st = i, ed = Math.min(chars.length - 1, i + k - 1); st < ed; ++st, --ed) {
-                char t = chars[st];
-                chars[st] = chars[ed];
-                chars[ed] = t;
+        char[] cs = s.toCharArray();
+        int n = cs.length;
+        for (int i = 0; i < n; i += k * 2) {
+            for (int l = i, r = Math.min(i + k - 1, n - 1); l < r; ++l, --r) {
+                char t = cs[l];
+                cs[l] = cs[r];
+                cs[r] = t;
             }
         }
-        return new String(chars);
+        return new String(cs);
     }
 }

solution/0500-0599/0541.Reverse String II/Solution.py

@@ -1,6 +1,6 @@
 class Solution:
     def reverseStr(self, s: str, k: int) -> str:
-        t = list(s)
-        for i in range(0, len(t), k << 1):
-            t[i : i + k] = reversed(t[i : i + k])
-        return ''.join(t)
+        cs = list(s)
+        for i in range(0, len(cs), 2 * k):
+            cs[i : i + k] = reversed(cs[i : i + k])
+        return "".join(cs)

solution/0500-0599/0541.Reverse String II/Solution.ts

@@ -0,0 +1,10 @@
+function reverseStr(s: string, k: number): string {
+    const n = s.length;
+    const cs = s.split('');
+    for (let i = 0; i < n; i += 2 * k) {
+        for (let l = i, r = Math.min(i + k - 1, n - 1); l < r; l++, r--) {
+            [cs[l], cs[r]] = [cs[r], cs[l]];
+        }
+    }
+    return cs.join('');
+}

solution/0500-0599/0542.01 Matrix/README.md

@@ -62,11 +62,17 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一：多源 BFS
+### 方法一：BFS
 
-初始化结果矩阵 ans，所有 0 的距离为 0，所以 1 的距离为 -1。初始化队列 q 存储 BFS 需要检查的位置，并将所有 0 的位置入队。
+我们创建一个大小和 $\textit{mat}$ 一样的矩阵 $\textit{ans}$，并将所有的元素初始化为 $-1$。
 
-循环弹出队列 q 的元素 `p(i, j)`，检查邻居四个点。对于邻居 `(x, y)`，如果 `ans[x][y] = -1`，则更新 `ans[x][y] = ans[i][j] + 1`。同时将 `(x, y)` 入队。
+然后我们遍历 $\textit{mat}$，将所有的 $0$ 元素的坐标 $(i, j)$ 加入队列 $\textit{q}$，并将 $\textit{ans}[i][j]$ 设为 $0$。
+
+接下来，我们使用广度优先搜索，从队列中取出一个元素 $(i, j)$，并遍历其四个方向，如果该方向的元素 $(x, y)$ 满足 $0 \leq x < m$, $0 \leq y < n$ 且 $\textit{ans}[x][y] = -1$，则将 $\textit{ans}[x][y]$ 设为 $\textit{ans}[i][j] + 1$，并将 $(x, y)$ 加入队列 $\textit{q}$。
+
+最后返回 $\textit{ans}$。
+
+时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别为矩阵 $\textit{mat}$ 的行数和列数。
 
 <!-- tabs:start -->
 
@@ -219,8 +225,7 @@ function updateMatrix(mat: number[][]): number[][] {
         }
     }
     const dirs: number[] = [-1, 0, 1, 0, -1];
-    while (q.length) {
-        const [i, j] = q.shift()!;
+    for (const [i, j] of q) {
         for (let k = 0; k < 4; ++k) {
             const [x, y] = [i + dirs[k], j + dirs[k + 1]];
             if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] === -1) {
@@ -239,49 +244,38 @@ function updateMatrix(mat: number[][]): number[][] {
 use std::collections::VecDeque;
 
 impl Solution {
-    #[allow(dead_code)]
     pub fn update_matrix(mat: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
-        let n: usize = mat.len();
-        let m: usize = mat[0].len();
-        let mut ret_vec: Vec<Vec<i32>> = vec![vec![-1; m]; n];
-        // The inner tuple is of <X, Y, Current Count>
-        let mut the_q: VecDeque<(usize, usize)> = VecDeque::new();
-        let traverse_vec: Vec<(i32, i32)> = vec![(-1, 0), (1, 0), (0, 1), (0, -1)];
-
-        // Initialize the queue
-        for i in 0..n {
-            for j in 0..m {
+        let m = mat.len();
+        let n = mat[0].len();
+        let mut ans = vec![vec![-1; n]; m];
+        let mut q = VecDeque::new();
+
+        for i in 0..m {
+            for j in 0..n {
                 if mat[i][j] == 0 {
-                    // For the zero cell, enqueue at first
-                    the_q.push_back((i, j));
-                    // Set to 0 in return vector
-                    ret_vec[i][j] = 0;
+                    q.push_back((i, j));
+                    ans[i][j] = 0;
                 }
             }
         }
 
-        while !the_q.is_empty() {
-            let (x, y) = the_q.front().unwrap().clone();
-            the_q.pop_front();
-            for pair in &traverse_vec {
-                let cur_x = pair.0 + (x as i32);
-                let cur_y = pair.1 + (y as i32);
-                if Solution::check_bounds(cur_x, cur_y, n as i32, m as i32)
-                    && ret_vec[cur_x as usize][cur_y as usize] == -1
-                {
-                    // The current cell has not be updated yet, and is also in bound
-                    ret_vec[cur_x as usize][cur_y as usize] = ret_vec[x][y] + 1;
-                    the_q.push_back((cur_x as usize, cur_y as usize));
+        let dirs = [-1, 0, 1, 0, -1];
+        while let Some((i, j)) = q.pop_front() {
+            for k in 0..4 {
+                let x = i as isize + dirs[k];
+                let y = j as isize + dirs[k + 1];
+                if x >= 0 && x < m as isize && y >= 0 && y < n as isize {
+                    let x = x as usize;
+                    let y = y as usize;
+                    if ans[x][y] == -1 {
+                        ans[x][y] = ans[i][j] + 1;
+                        q.push_back((x, y));
+                    }
                 }
             }
         }
 
-        ret_vec
-    }
-
-    #[allow(dead_code)]
-    pub fn check_bounds(i: i32, j: i32, n: i32, m: i32) -> bool {
-        i >= 0 && i < n && j >= 0 && j < m
+        ans
     }
 }
 ```