feat: add solutions to lc problem: No.240 (#3898)

No.0240.Search a 2D Matrix II

solution/0000-0099/0048.Rotate Image/README.md

@@ -58,9 +58,9 @@ tags:
 
 ### 方法一：原地翻转
 
-根据题目要求，我们实际上需要将 $matrix[i][j]$ 旋转至 $matrix[j][n - i - 1]$。
+根据题目要求，我们实际上需要将 $\text{matrix}[i][j]$ 旋转至 $\text{matrix}[j][n - i - 1]$。
 
-我们可以先对矩阵进行上下翻转，即 $matrix[i][j]$ 和 $matrix[n - i - 1][j]$ 进行交换，然后再对矩阵进行主对角线翻转，即 $matrix[i][j]$ 和 $matrix[j][i]$ 进行交换。这样就能将 $matrix[i][j]$ 旋转至 $matrix[j][n - i - 1]$ 了。
+我们可以先对矩阵进行上下翻转，即 $\text{matrix}[i][j]$ 和 $\text{matrix}[n - i - 1][j]$ 进行交换，然后再对矩阵进行主对角线翻转，即 $\text{matrix}[i][j]$ 和 $\text{matrix}[j][i]$ 进行交换。这样就能将 $\text{matrix}[i][j]$ 旋转至 $\text{matrix}[j][n - i - 1]$ 了。
 
 时间复杂度 $O(n^2)$，其中 $n$ 是矩阵的边长。空间复杂度 $O(1)$。
 

solution/0000-0099/0048.Rotate Image/README_EN.md

@@ -54,9 +54,9 @@ tags:
 
 ### Solution 1: In-place Rotation
 
-According to the problem requirements, we actually need to rotate $matrix[i][j]$ to $matrix[j][n - i - 1]$.
+According to the problem requirements, we need to rotate $\text{matrix}[i][j]$ to $\text{matrix}[j][n - i - 1]$.
 
-We can first flip the matrix upside down, that is, swap $matrix[i][j]$ and $matrix[n - i - 1][j]$, and then flip the matrix along the main diagonal, that is, swap $matrix[i][j]$ and $matrix[j][i]$. This way, we can rotate $matrix[i][j]$ to $matrix[j][n - i - 1]$.
+We can first flip the matrix upside down, i.e., swap $\text{matrix}[i][j]$ with $\text{matrix}[n - i - 1][j]$, and then flip the matrix along the main diagonal, i.e., swap $\text{matrix}[i][j]$ with $\text{matrix}[j][i]$. This way, we can rotate $\text{matrix}[i][j]$ to $\text{matrix}[j][n - i - 1]$.
 
 The time complexity is $O(n^2)$, where $n$ is the side length of the matrix. The space complexity is $O(1)$.
 

solution/0200-0299/0240.Search a 2D Matrix II/README.md

@@ -64,9 +64,9 @@ tags:
 
 ### 方法一：二分查找
 
-由于每一行的所有元素升序排列，因此，对于每一行，我们可以使用二分查找找到第一个大于等于 `target` 的元素，然后判断该元素是否等于 `target`。如果等于 `target`，说明找到了目标值，直接返回 `true`。如果不等于 `target`，说明这一行的所有元素都小于 `target`，应该继续搜索下一行。
+由于每一行的所有元素升序排列，因此，对于每一行，我们可以使用二分查找找到第一个大于等于 $\textit{target}$ 的元素，然后判断该元素是否等于 $\textit{target}$。如果等于 $\textit{target}$，说明找到了目标值，直接返回 $\text{true}$。如果不等于 $\textit{target}$，说明这一行的所有元素都小于 $\textit{target}$，应该继续搜索下一行。
 
-如果所有行都搜索完了，都没有找到目标值，说明目标值不存在，返回 `false`。
+如果所有行都搜索完了，都没有找到目标值，说明目标值不存在，返回 $\text{false}$。
 
 时间复杂度 $O(m \times \log n)$，其中 $m$ 和 $n$ 分别为矩阵的行数和列数。空间复杂度 $O(1)$。
 
@@ -137,17 +137,8 @@ func searchMatrix(matrix [][]int, target int) bool {
 function searchMatrix(matrix: number[][], target: number): boolean {
     const n = matrix[0].length;
     for (const row of matrix) {
-        let left = 0,
-            right = n;
-        while (left < right) {
-            const mid = (left + right) >> 1;
-            if (row[mid] >= target) {
-                right = mid;
-            } else {
-                left = mid + 1;
-            }
-        }
-        if (left != n && row[left] == target) {
+        const j = _.sortedIndex(row, target);
+        if (j < n && row[j] === target) {
             return true;
         }
     }
@@ -195,17 +186,8 @@ impl Solution {
 var searchMatrix = function (matrix, target) {
     const n = matrix[0].length;
     for (const row of matrix) {
-        let left = 0,
-            right = n;
-        while (left < right) {
-            const mid = (left + right) >> 1;
-            if (row[mid] >= target) {
-                right = mid;
-            } else {
-                left = mid + 1;
-            }
-        }
-        if (left != n && row[left] == target) {
+        const j = _.sortedIndex(row, target);
+        if (j < n && row[j] == target) {
             return true;
         }
     }
@@ -237,13 +219,13 @@ public class Solution {
 
 ### 方法二：从左下角或右上角搜索
 
-这里我们以左下角作为起始搜索点，往右上方向开始搜索，比较当前元素 `matrix[i][j]`与 `target` 的大小关系：
+这里我们以左下角或右上角作为起始搜索点，往右上或左下方向开始搜索。比较当前元素 $\textit{matrix}[i][j]$ 与 $\textit{target}$ 的大小关系：
 
--   若 $\textit{matrix}[i][j] = \textit{target}$，说明找到了目标值，直接返回 `true`。
--   若 $\textit{matrix}[i][j] > \textit{target}$，说明这一列从当前位置开始往上的所有元素均大于 `target`，应该让 $i$ 指针往上移动，即 $i \leftarrow i - 1$。
--   若 $\textit{matrix}[i][j] < \textit{target}$，说明这一行从当前位置开始往右的所有元素均小于 `target`，应该让 $j$ 指针往右移动，即 $j \leftarrow j + 1$。
+-   若 $\textit{matrix}[i][j] = \textit{target}$，说明找到了目标值，直接返回 $\text{true}$。
+-   若 $\textit{matrix}[i][j] > \textit{target}$，说明这一列从当前位置开始往上的所有元素均大于 $\textit{target}$，应该让 $i$ 指针往上移动，即 $i \leftarrow i - 1$。
+-   若 $\textit{matrix}[i][j] < \textit{target}$，说明这一行从当前位置开始往右的所有元素均小于 $\textit{target}$，应该让 $j$ 指针往右移动，即 $j \leftarrow j + 1$。
 
-若搜索结束依然找不到 `target`，返回 `false`。
+若搜索结束依然找不到 $\textit{target}$，返回 $\text{false}$。
 
 时间复杂度 $O(m + n)$，其中 $m$ 和 $n$ 分别为矩阵的行数和列数。空间复杂度 $O(1)$。
 
@@ -351,6 +333,33 @@ function searchMatrix(matrix: number[][], target: number): boolean {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn search_matrix(matrix: Vec<Vec<i32>>, target: i32) -> bool {
+        let m = matrix.len();
+        let n = matrix[0].len();
+        let mut i = m - 1;
+        let mut j = 0;
+        while i >= 0 && j < n {
+            if matrix[i][j] == target {
+                return true;
+            }
+            if matrix[i][j] > target {
+                if i == 0 {
+                    break;
+                }
+                i -= 1;
+            } else {
+                j += 1;
+            }
+        }
+        false
+    }
+}
+```
+
 #### C#
 
 ```cs

solution/0200-0299/0240.Search a 2D Matrix II/README_EN.md

@@ -62,11 +62,11 @@ tags:
 
 ### Solution 1: Binary Search
 
-Since all elements in each row are sorted in ascending order, we can use binary search to find the first element that is greater than or equal to `target` for each row, and then check if this element is equal to `target`. If it equals `target`, it means the target value has been found, and we directly return `true`. If it does not equal `target`, it means all elements in this row are less than `target`, and we should continue to search the next row.
+Since all elements in each row are sorted in ascending order, for each row, we can use binary search to find the first element greater than or equal to $\textit{target}$, and then check if that element is equal to $\textit{target}$. If it is equal to $\textit{target}$, it means the target value is found, and we return $\text{true}$. If it is not equal to $\textit{target}$, it means all elements in this row are less than $\textit{target}$, and we should continue searching the next row.
 
-If all rows have been searched and the target value has not been found, it means the target value does not exist, so we return `false`.
+If all rows have been searched and the target value is not found, it means the target value does not exist, and we return $\text{false}$.
 
-The time complexity is $O(m \times \log n)$, where $m$ and $n$ are the number of rows and columns in the matrix, respectively. The space complexity is $O(1)$.
+The time complexity is $O(m \times \log n)$, where $m$ and $n$ are the number of rows and columns of the matrix, respectively. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -135,17 +135,8 @@ func searchMatrix(matrix [][]int, target int) bool {
 function searchMatrix(matrix: number[][], target: number): boolean {
     const n = matrix[0].length;
     for (const row of matrix) {
-        let left = 0,
-            right = n;
-        while (left < right) {
-            const mid = (left + right) >> 1;
-            if (row[mid] >= target) {
-                right = mid;
-            } else {
-                left = mid + 1;
-            }
-        }
-        if (left != n && row[left] == target) {
+        const j = _.sortedIndex(row, target);
+        if (j < n && row[j] === target) {
             return true;
         }
     }
@@ -193,17 +184,8 @@ impl Solution {
 var searchMatrix = function (matrix, target) {
     const n = matrix[0].length;
     for (const row of matrix) {
-        let left = 0,
-            right = n;
-        while (left < right) {
-            const mid = (left + right) >> 1;
-            if (row[mid] >= target) {
-                right = mid;
-            } else {
-                left = mid + 1;
-            }
-        }
-        if (left != n && row[left] == target) {
+        const j = _.sortedIndex(row, target);
+        if (j < n && row[j] == target) {
             return true;
         }
     }
@@ -233,17 +215,17 @@ public class Solution {
 
 <!-- solution:start -->
 
-### Solution 2: Search from the Bottom Left or Top Right
+### Solution 2: Search from Bottom-Left or Top-Right
 
-Here, we start searching from the bottom left corner and move towards the top right direction, comparing the current element `matrix[i][j]` with `target`:
+We start the search from the bottom-left or top-right corner and move towards the top-right or bottom-left direction. Compare the current element $\textit{matrix}[i][j]$ with $\textit{target}$:
 
--   If $\textit{matrix}[i][j] = \textit{target}$, it means the target value has been found, and we directly return `true`.
--   If $\textit{matrix}[i][j] > \textit{target}$, it means all elements in this column from the current position upwards are greater than `target`, so we should move the $i$ pointer upwards, i.e., $i \leftarrow i - 1$.
--   If $\textit{matrix}[i][j] < \textit{target}$, it means all elements in this row from the current position to the right are less than `target`, so we should move the $j$ pointer to the right, i.e., $j \leftarrow j + 1$.
+-   If $\textit{matrix}[i][j] = \textit{target}$, it means the target value is found, and we return $\text{true}$.
+-   If $\textit{matrix}[i][j] > \textit{target}$, it means all elements in this column from the current position upwards are greater than $\textit{target}$, so we move the $i$ pointer upwards, i.e., $i \leftarrow i - 1$.
+-   If $\textit{matrix}[i][j] < \textit{target}$, it means all elements in this row from the current position to the right are less than $\textit{target}$, so we move the $j$ pointer to the right, i.e., $j \leftarrow j + 1$.
 
-If the search ends and the `target` is still not found, return `false`.
+If the search ends and the $\textit{target}$ is not found, return $\text{false}$.
 
-The time complexity is $O(m + n)$, where $m$ and $n$ are the number of rows and columns in the matrix, respectively. The space complexity is $O(1)$.
+The time complexity is $O(m + n)$, where $m$ and $n$ are the number of rows and columns of the matrix, respectively. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -349,6 +331,33 @@ function searchMatrix(matrix: number[][], target: number): boolean {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn search_matrix(matrix: Vec<Vec<i32>>, target: i32) -> bool {
+        let m = matrix.len();
+        let n = matrix[0].len();
+        let mut i = m - 1;
+        let mut j = 0;
+        while i >= 0 && j < n {
+            if matrix[i][j] == target {
+                return true;
+            }
+            if matrix[i][j] > target {
+                if i == 0 {
+                    break;
+                }
+                i -= 1;
+            } else {
+                j += 1;
+            }
+        }
+        false
+    }
+}
+```
+
 #### C#
 
 ```cs

solution/0200-0299/0240.Search a 2D Matrix II/Solution.js

@@ -6,17 +6,8 @@
 var searchMatrix = function (matrix, target) {
     const n = matrix[0].length;
     for (const row of matrix) {
-        let left = 0,
-            right = n;
-        while (left < right) {
-            const mid = (left + right) >> 1;
-            if (row[mid] >= target) {
-                right = mid;
-            } else {
-                left = mid + 1;
-            }
-        }
-        if (left != n && row[left] == target) {
+        const j = _.sortedIndex(row, target);
+        if (j < n && row[j] == target) {
             return true;
         }
     }

solution/0200-0299/0240.Search a 2D Matrix II/Solution.ts

@@ -1,17 +1,8 @@
 function searchMatrix(matrix: number[][], target: number): boolean {
     const n = matrix[0].length;
     for (const row of matrix) {
-        let left = 0,
-            right = n;
-        while (left < right) {
-            const mid = (left + right) >> 1;
-            if (row[mid] >= target) {
-                right = mid;
-            } else {
-                left = mid + 1;
-            }
-        }
-        if (left != n && row[left] == target) {
+        const j = _.sortedIndex(row, target);
+        if (j < n && row[j] === target) {
             return true;
         }
     }

solution/0200-0299/0240.Search a 2D Matrix II/Solution2.rs

@@ -0,0 +1,22 @@
+impl Solution {
+    pub fn search_matrix(matrix: Vec<Vec<i32>>, target: i32) -> bool {
+        let m = matrix.len();
+        let n = matrix[0].len();
+        let mut i = m - 1;
+        let mut j = 0;
+        while i >= 0 && j < n {
+            if matrix[i][j] == target {
+                return true;
+            }
+            if matrix[i][j] > target {
+                if i == 0 {
+                    break;
+                }
+                i -= 1;
+            } else {
+                j += 1;
+            }
+        }
+        false
+    }
+}