feat: add solutions to lc problem: No.0228

No.0228.Summary Ranges

Author: yanglbme

solution/0200-0299/0228.Summary Ranges/README.md

@@ -63,7 +63,7 @@
 
 遍历数组，当 $j + 1 < n$ 且 $nums[j + 1] = nums[j] + 1$ 时，$j$ 向右移动，否则区间 $[i, j]$ 已经找到，将其加入答案，然后将 $i$ 移动到 $j + 1$ 的位置，继续寻找下一个区间。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
+时间复杂度 $O(n)$，其中 $n$ 为数组长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -74,7 +74,7 @@
 ```python
 class Solution:
     def summaryRanges(self, nums: List[int]) -> List[str]:
-        def f(i, j):
+        def f(i: int, j: int) -> str:
             return str(nums[i]) if i == j else f'{nums[i]}->{nums[j]}'
 
         i = 0
@@ -156,6 +156,26 @@ func summaryRanges(nums []int) (ans []string) {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function summaryRanges(nums: number[]): string[] {
+    const f = (i: number, j: number): string => {
+        return i === j ? `${nums[i]}` : `${nums[i]}->${nums[j]}`;
+    };
+    const n = nums.length;
+    const ans: string[] = [];
+    for (let i = 0, j = 0; i < n; i = j + 1) {
+        j = i;
+        while (j + 1 < n && nums[j + 1] === nums[j] + 1) {
+            ++j;
+        }
+        ans.push(f(i, j));
+    }
+    return ans;
+}
+```
+
 ### **C#**
 
 ```cs

solution/0200-0299/0228.Summary Ranges/README_EN.md

@@ -53,14 +53,22 @@
 
 ## Solutions
 
+**Approach 1: Two Pointers**
+
+We can use two pointers $i$ and $j$ to find the left and right endpoints of each interval.
+
+Traverse the array, when $j + 1 < n$ and $nums[j + 1] = nums[j] + 1$, move $j$ to the right, otherwise the interval $[i, j]$ has been found, add it to the answer, then move $i$ to the position of $j + 1$, and continue to find the next interval.
+
+Time complexity $O(n)$, where $n$ is the length of the array. Space complexity $O(1)$.
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
 class Solution:
     def summaryRanges(self, nums: List[int]) -> List[str]:
-        def f(i, j):
+        def f(i: int, j: int) -> str:
             return str(nums[i]) if i == j else f'{nums[i]}->{nums[j]}'
 
         i = 0
@@ -140,6 +148,26 @@ func summaryRanges(nums []int) (ans []string) {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function summaryRanges(nums: number[]): string[] {
+    const f = (i: number, j: number): string => {
+        return i === j ? `${nums[i]}` : `${nums[i]}->${nums[j]}`;
+    };
+    const n = nums.length;
+    const ans: string[] = [];
+    for (let i = 0, j = 0; i < n; i = j + 1) {
+        j = i;
+        while (j + 1 < n && nums[j + 1] === nums[j] + 1) {
+            ++j;
+        }
+        ans.push(f(i, j));
+    }
+    return ans;
+}
+```
+
 ### **C#**
 
 ```cs

solution/0200-0299/0228.Summary Ranges/Solution.py

@@ -1,6 +1,6 @@
 class Solution:
     def summaryRanges(self, nums: List[int]) -> List[str]:
-        def f(i, j):
+        def f(i: int, j: int) -> str:
             return str(nums[i]) if i == j else f'{nums[i]}->{nums[j]}'
 
         i = 0

solution/0200-0299/0228.Summary Ranges/Solution.ts

@@ -0,0 +1,15 @@
+function summaryRanges(nums: number[]): string[] {
+    const f = (i: number, j: number): string => {
+        return i === j ? `${nums[i]}` : `${nums[i]}->${nums[j]}`;
+    };
+    const n = nums.length;
+    const ans: string[] = [];
+    for (let i = 0, j = 0; i < n; i = j + 1) {
+        j = i;
+        while (j + 1 < n && nums[j + 1] === nums[j] + 1) {
+            ++j;
+        }
+        ans.push(f(i, j));
+    }
+    return ans;
+}