feat: add cs solution to lc problem: No.0935 (#2023)

Author: dev-mauli

solution/0900-0999/0935.Knight Dialer/README.md

@@ -195,6 +195,35 @@ func knightDialer(n int) int {
 }
 ```
 
+### **C#**
+
+```cs
+public class Solution {
+    public int KnightDialer(int n) {
+        if (n == 1) return 10;
+        int A = 4;
+        int B = 2;
+        int C = 2;
+        int D = 1;
+        int MOD = (int)1e9 + 7;
+        for (int i = 0; i < n - 1; i++) {
+            int tempA = A;
+            int tempB = B;
+            int tempC = C;
+            int tempD = D;
+            A = ((2 * tempB) % MOD + (2 * tempC) % MOD) % MOD;
+            B = tempA;
+            C = (tempA + (2 * tempD) % MOD) % MOD;
+            D = tempC;
+        }
+
+        int ans = (A + B) % MOD;
+        ans = (ans + C) % MOD;
+        return (ans + D) % MOD;
+    }
+}
+```
+
 ### **...**
 
 ```

solution/0900-0999/0935.Knight Dialer/README_EN.md

@@ -176,6 +176,35 @@ func knightDialer(n int) int {
 }
 ```
 
+### **C#**
+
+```cs
+public class Solution {
+    public int KnightDialer(int n) {
+        if (n == 1) return 10;
+        int A = 4;
+        int B = 2;
+        int C = 2;
+        int D = 1;
+        int MOD = (int)1e9 + 7;
+        for (int i = 0; i < n - 1; i++) {
+            int tempA = A;
+            int tempB = B;
+            int tempC = C;
+            int tempD = D;
+            A = ((2 * tempB) % MOD + (2 * tempC) % MOD) % MOD;
+            B = tempA;
+            C = (tempA + (2 * tempD) % MOD) % MOD;
+            D = tempC;
+        }
+
+        int ans = (A + B) % MOD;
+        ans = (ans + C) % MOD;
+        return (ans + D) % MOD;
+    }
+}
+```
+
 ### **...**
 
 ```

solution/0900-0999/0935.Knight Dialer/Solution.cs

@@ -0,0 +1,24 @@
+public class Solution {
+    public int KnightDialer(int n) {
+        if (n == 1) return 10;
+        int A = 4;
+        int B = 2;
+        int C = 2;
+        int D = 1;
+        int MOD = (int)1e9 + 7;
+        for (int i = 0; i < n - 1; i++) {
+            int tempA = A;
+            int tempB = B;
+            int tempC = C;
+            int tempD = D;
+            A = ((2 * tempB) % MOD + (2 * tempC) % MOD) % MOD;
+            B = tempA;
+            C = (tempA + (2 * tempD) % MOD) % MOD;
+            D = tempC;
+        }
+        
+        int ans = (A + B) % MOD;
+        ans = (ans + C) % MOD;
+        return (ans + D) % MOD;
+    }
+}