feat: add typescript solution to lc problem: No.1888.Minimum Number of Flips to Make the Binary String Alternating (#441)

Author: zhaocchen

solution/1800-1899/1888.Minimum Number of Flips to Make the Binary String Alternating/README.md

@@ -108,6 +108,26 @@ class Solution {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function minFlips(s: string): number {
+    const n: number = s.length;
+    const target: string[] = ['0', '1'];
+    let count: number = 0;
+    for (let i: number = 0; i < n; ++i) {
+        count += (s.charAt(i) == target[i & 1] ? 0 : 1);
+    }
+    let res = Math.min(count, n - count);
+    for (let i: number = 0; i < n; ++i) {
+        count -= (s.charAt(i) == target[i & 1] ? 0 : 1);
+        count += (s.charAt(i) == target[(i + n) & 1] ? 0 : 1);
+        res = Math.min(res, count, n - count);
+    }
+    return res;
+};
+```
+
 ### **...**
 
 ```

solution/1800-1899/1888.Minimum Number of Flips to Make the Binary String Alternating/README_EN.md

@@ -100,6 +100,26 @@ class Solution {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function minFlips(s: string): number {
+    const n: number = s.length;
+    const target: string[] = ['0', '1'];
+    let count: number = 0;
+    for (let i: number = 0; i < n; ++i) {
+        count += (s.charAt(i) == target[i & 1] ? 0 : 1);
+    }
+    let res = Math.min(count, n - count);
+    for (let i: number = 0; i < n; ++i) {
+        count -= (s.charAt(i) == target[i & 1] ? 0 : 1);
+        count += (s.charAt(i) == target[(i + n) & 1] ? 0 : 1);
+        res = Math.min(res, count, n - count);
+    }
+    return res;
+};
+```
+
 ### **...**
 
 ```

solution/1800-1899/1888.Minimum Number of Flips to Make the Binary String Alternating/Solution.ts

@@ -0,0 +1,15 @@
+function minFlips(s: string): number {
+    const n: number = s.length;
+    const target: string[] = ['0', '1'];
+    let count: number = 0;
+    for (let i: number = 0; i < n; ++i) {
+        count += (s.charAt(i) == target[i & 1] ? 0 : 1);
+    }
+    let res = Math.min(count, n - count);
+    for (let i: number = 0; i < n; ++i) {
+        count -= (s.charAt(i) == target[i & 1] ? 0 : 1);
+        count += (s.charAt(i) == target[(i + n) & 1] ? 0 : 1);
+        res = Math.min(res, count, n - count);
+    }
+    return res;
+};