feat: add typescript solution to lc problem: No.0463.Island Perimeter (doocs#477)

Author: zhaocchen

solution/0400-0499/0463.Island Perimeter/README.md

@@ -53,6 +53,8 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+遍历二维数组
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -71,6 +73,37 @@
 
 ```
 
+### **TypeScript**
+
+```ts
+function islandPerimeter(grid: number[][]): number {
+    let m = grid.length, n = grid[0].length;
+    let ans = 0;
+    for (let i = 0; i < m; ++i) {
+        for (let j = 0; j < n; ++j) {
+            let top = 0, left = 0;
+            if (i > 0) {
+                top = grid[i - 1][j];
+            }
+            if (j > 0) {
+                left = grid[i][j - 1];
+            }
+            let cur = grid[i][j];
+            if (cur != top) ++ans;
+            if (cur != left) ++ans;
+        }
+    }
+    // 最后一行， 最后一列
+    for (let i = 0; i < m; ++i) {
+        if (grid[i][n - 1] == 1) ++ans;
+    }
+    for (let j = 0; j < n; ++j) {
+        if (grid[m - 1][j] == 1) ++ans;
+    }
+    return ans;
+};
+```
+
 ### **...**
 
 ```

solution/0400-0499/0463.Island Perimeter/README_EN.md

@@ -60,6 +60,37 @@
 
 ```
 
+### **TypeScript**
+
+```ts
+function islandPerimeter(grid: number[][]): number {
+    let m = grid.length, n = grid[0].length;
+    let ans = 0;
+    for (let i = 0; i < m; ++i) {
+        for (let j = 0; j < n; ++j) {
+            let top = 0, left = 0;
+            if (i > 0) {
+                top = grid[i - 1][j];
+            }
+            if (j > 0) {
+                left = grid[i][j - 1];
+            }
+            let cur = grid[i][j];
+            if (cur != top) ++ans;
+            if (cur != left) ++ans;
+        }
+    }
+    // 最后一行， 最后一列
+    for (let i = 0; i < m; ++i) {
+        if (grid[i][n - 1] == 1) ++ans;
+    }
+    for (let j = 0; j < n; ++j) {
+        if (grid[m - 1][j] == 1) ++ans;
+    }
+    return ans;
+};
+```
+
 ### **...**
 
 ```

solution/0400-0499/0463.Island Perimeter/Solution.ts

@@ -0,0 +1,26 @@
+function islandPerimeter(grid: number[][]): number {
+    let m = grid.length, n = grid[0].length;
+    let ans = 0;
+    for (let i = 0; i < m; ++i) {
+        for (let j = 0; j < n; ++j) {
+            let top = 0, left = 0;
+            if (i > 0) {
+                top = grid[i - 1][j];
+            }
+            if (j > 0) {
+                left = grid[i][j - 1];
+            }
+            let cur = grid[i][j];
+            if (cur != top) ++ans;
+            if (cur != left) ++ans;
+        }
+    }
+    // 最后一行， 最后一列
+    for (let i = 0; i < m; ++i) {
+        if (grid[i][n - 1] == 1) ++ans;
+    }
+    for (let j = 0; j < n; ++j) {
+        if (grid[m - 1][j] == 1) ++ans;
+    }
+    return ans;
+};