feat: add solutions to lc problems: No.1040

Author: thinkasany

solution/1000-1099/1040.Moving Stones Until Consecutive II/README.md

@@ -176,6 +176,30 @@ func min(a, b int) int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function numMovesStonesII(stones: number[]): number[] {
+    stones.sort((a, b) => a - b);
+    const n = stones.length;
+    let mi = n;
+    const mx =
+        Math.max(stones[n - 1] - stones[1] + 1, stones[n - 2] - stones[0] + 1) -
+        (n - 1);
+    for (let i = 0, j = 0; j < n; ++j) {
+        while (stones[j] - stones[i] + 1 > n) {
+            ++i;
+        }
+        if (j - i + 1 === n - 1 && stones[j] - stones[i] === n - 2) {
+            mi = Math.min(mi, 2);
+        } else {
+            mi = Math.min(mi, n - (j - i + 1));
+        }
+    }
+    return [mi, mx];
+}
+```
+
 ### **...**
 
 ```

solution/1000-1099/1040.Moving Stones Until Consecutive II/README_EN.md

@@ -159,6 +159,30 @@ func min(a, b int) int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function numMovesStonesII(stones: number[]): number[] {
+    stones.sort((a, b) => a - b);
+    const n = stones.length;
+    let mi = n;
+    const mx =
+        Math.max(stones[n - 1] - stones[1] + 1, stones[n - 2] - stones[0] + 1) -
+        (n - 1);
+    for (let i = 0, j = 0; j < n; ++j) {
+        while (stones[j] - stones[i] + 1 > n) {
+            ++i;
+        }
+        if (j - i + 1 === n - 1 && stones[j] - stones[i] === n - 2) {
+            mi = Math.min(mi, 2);
+        } else {
+            mi = Math.min(mi, n - (j - i + 1));
+        }
+    }
+    return [mi, mx];
+}
+```
+
 ### **...**
 
 ```

solution/1000-1099/1040.Moving Stones Until Consecutive II/Solution.ts

@@ -0,0 +1,19 @@
+function numMovesStonesII(stones: number[]): number[] {
+    stones.sort((a, b) => a - b);
+    const n = stones.length;
+    let mi = n;
+    const mx =
+        Math.max(stones[n - 1] - stones[1] + 1, stones[n - 2] - stones[0] + 1) -
+        (n - 1);
+    for (let i = 0, j = 0; j < n; ++j) {
+        while (stones[j] - stones[i] + 1 > n) {
+            ++i;
+        }
+        if (j - i + 1 === n - 1 && stones[j] - stones[i] === n - 2) {
+            mi = Math.min(mi, 2);
+        } else {
+            mi = Math.min(mi, n - (j - i + 1));
+        }
+    }
+    return [mi, mx];
+}