feat: add typescript solution to lc problem: No.0064.Minimum Path Sum (doocs#531)

Author: zhaocchen

solution/0000-0099/0064.Minimum Path Sum/README.md

@@ -93,6 +93,30 @@ class Solution {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function minPathSum(grid: number[][]): number {
+   let m = grid.length, n = grid[0].length;
+   let dp = Array.from({ length: m}, v => new Array(n).fill(0));
+   dp[0][0] = grid[0][0];
+   for (let i = 1; i < m; ++i) {
+       dp[i][0] = dp[i - 1][0] + grid[i][0];
+   }
+   for (let j = 1; j < n; ++j) {
+       dp[0][j] = dp[0][j - 1] + grid[0][j];
+   }
+   // dp
+   for (let i = 1; i < m; ++i) {
+       for (let j = 1; j < n; ++j) {
+           let cur = grid[i][j];
+           dp[i][j] = cur + Math.min(dp[i - 1][j], dp[i][j - 1]);
+       }
+   }
+   return dp[m - 1][n - 1];
+};
+```
+
 ### **C++**
 
 ```cpp

solution/0000-0099/0064.Minimum Path Sum/README_EN.md

@@ -81,6 +81,30 @@ class Solution {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function minPathSum(grid: number[][]): number {
+   let m = grid.length, n = grid[0].length;
+   let dp = Array.from({ length: m}, v => new Array(n).fill(0));
+   dp[0][0] = grid[0][0];
+   for (let i = 1; i < m; ++i) {
+       dp[i][0] = dp[i - 1][0] + grid[i][0];
+   }
+   for (let j = 1; j < n; ++j) {
+       dp[0][j] = dp[0][j - 1] + grid[0][j];
+   }
+   // dp
+   for (let i = 1; i < m; ++i) {
+       for (let j = 1; j < n; ++j) {
+           let cur = grid[i][j];
+           dp[i][j] = cur + Math.min(dp[i - 1][j], dp[i][j - 1]);
+       }
+   }
+   return dp[m - 1][n - 1];
+};
+```
+
 ### **C++**
 
 ```cpp

solution/0000-0099/0064.Minimum Path Sum/Solution.ts

@@ -0,0 +1,19 @@
+function minPathSum(grid: number[][]): number {
+    let m = grid.length, n = grid[0].length;
+    let dp = Array.from({ length: m}, v => new Array(n).fill(0));
+    dp[0][0] = grid[0][0];
+    for (let i = 1; i < m; ++i) {
+        dp[i][0] = dp[i - 1][0] + grid[i][0];
+    }
+    for (let j = 1; j < n; ++j) {
+        dp[0][j] = dp[0][j - 1] + grid[0][j];
+    }
+    // dp
+    for (let i = 1; i < m; ++i) {
+        for (let j = 1; j < n; ++j) {
+            let cur = grid[i][j];
+            dp[i][j] = cur + Math.min(dp[i - 1][j], dp[i][j - 1]);
+        }
+    }
+    return dp[m - 1][n - 1];
+ };