feat: add No.959

Author: BaffinLee

901-1000/959. Regions Cut By Slashes.md

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+# 959. Regions Cut By Slashes
+
+- Difficulty: Medium.
+- Related Topics: Array, Hash Table, Depth-First Search, Breadth-First Search, Union Find, Matrix.
+- Similar Questions: .
+
+## Problem
+
+An `n x n` grid is composed of `1 x 1` squares where each `1 x 1` square consists of a `'/'`, `'\'`, or blank space `' '`. These characters divide the square into contiguous regions.
+
+Given the grid `grid` represented as a string array, return **the number of regions**.
+
+Note that backslash characters are escaped, so a `'\'` is represented as `'\\'`.
+
+ 
+Example 1:
+
+![](https://assets.leetcode.com/uploads/2018/12/15/1.png)
+
+```
+Input: grid = [" /","/ "]
+Output: 2
+```
+
+Example 2:
+
+![](https://assets.leetcode.com/uploads/2018/12/15/2.png)
+
+```
+Input: grid = [" /","  "]
+Output: 1
+```
+
+Example 3:
+
+![](https://assets.leetcode.com/uploads/2018/12/15/4.png)
+
+```
+Input: grid = ["/\\","\\/"]
+Output: 5
+Explanation: Recall that because \ characters are escaped, "\\/" refers to \/, and "/\\" refers to /\.
+```
+
+ 
+**Constraints:**
+
+
+	
+- `n == grid.length == grid[i].length`
+	
+- `1 <= n <= 30`
+	
+- `grid[i][j]` is either `'/'`, `'\'`, or `' '`.
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {string[]} grid
+ * @return {number}
+ */
+var regionsBySlashes = function(grid) {
+    var matrix = Array(3 * grid.length).fill(0).map(() => Array(3 * grid.length).fill(0));
+    for (var i = 0; i < grid.length; i++) {
+        for (var j = 0; j < grid[i].length; j++) {
+            if (grid[i][j] === '\\') {
+                matrix[i * 3][j * 3] = 1;
+                matrix[i * 3 + 1][j * 3 + 1] = 1;
+                matrix[i * 3 + 2][j * 3 + 2] = 1;
+            } else if (grid[i][j] === '/') {
+                matrix[i * 3][j * 3 + 2] = 1;
+                matrix[i * 3 + 1][j * 3 + 1] = 1;
+                matrix[i * 3 + 2][j * 3] = 1;
+            }
+        }
+    }
+    var res = 0;
+    for (var i = 0; i < matrix.length; i++) {
+        for (var j = 0; j < matrix[i].length; j++) {
+            if (matrix[i][j] === 0) {
+                visitAndMark(matrix, i, j, ++res);
+            }
+        }
+    }
+    return res;
+};
+
+var visitAndMark = function(matrix, i, j, num) {
+    matrix[i][j] = num;
+    matrix[i][j + 1] === 0 && visitAndMark(matrix, i, j + 1, num);
+    matrix[i][j - 1] === 0 && visitAndMark(matrix, i, j - 1, num);
+    matrix[i + 1]?.[j] === 0 && visitAndMark(matrix, i + 1, j, num);
+    matrix[i - 1]?.[j] === 0 && visitAndMark(matrix, i - 1, j, num);
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n ^ 2).
+* Space complexity : O(n ^ 2).