doocs · yanglbme · Jun 22, 2021 · Jun 22, 2021
diff --git a/README_EN.md b/README_EN.md
@@ -43,7 +43,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
 
 ### Searching
 
-- [Binary Search(Algorithm Template)](./basic/searching/BinarySearch/README.md)
+- [Binary Search(Algorithm Template)](./basic/searching/BinarySearch/README_EN.md)
 
 ## High Frequency Interview Questions
 

diff --git a/basic/searching/BinarySearch/Main.java b/basic/searching/BinarySearch/Main.java
diff --git a/basic/searching/BinarySearch/Main.py b/basic/searching/BinarySearch/Main.py
diff --git a/basic/searching/BinarySearch/README.md b/basic/searching/BinarySearch/README.md
@@ -29,128 +29,6 @@ int binarySearch2(int left, int right) {
 }
 ```
 
-## 题目描述
+## 例题
 
-给定一个按照升序排列的长度为 `n` 的整数数组，以及 `q` 个查询。
-
-对于每个查询，返回一个元素 k 的起始位置和终止位置（位置从 0 开始计数）。
-
-如果数组中不存在该元素，则返回 `-1 -1`。
-
-**输入格式**
-
-第一行包含整数 n 和 q，表示数组长度和询问个数。
-
-第二行包含  n  个整数（均在 1∼10000 范围内），表示完整数组。
-
-接下来 q 行，每行包含一个整数 k，表示一个询问元素。
-
-**输出格式**
-
-共 q 行，每行包含两个整数，表示所求元素的起始位置和终止位置。
-
-如果数组中不存在该元素，则返回 `-1 -1`。
-
-**数据范围**
-
-- 1≤n≤100000
-- 1≤q≤10000
-- 1≤k≤10000
-
-**输入样例：**
-
-```
-6 3
-1 2 2 3 3 4
-3
-4
-5
-```
-
-**输出样例：**
-
-```
-3 4
-5 5
--1 -1
-```
-
-## 代码实现
-
-<!-- tabs:start -->
-
-### **Python3**
-
-```python
-n, q = map(int, input().split())
-nums = list(map(int, input().split()))
-
-for _ in range(q):
-    x = int(input())
-    left, right = 0, n - 1
-    while left < right:
-        mid = (left + right) >> 1
-        if nums[mid] >= x:
-            right = mid
-        else:
-            left = mid + 1
-    if nums[left] != x:
-        print('-1 -1')
-    else:
-        t = left
-        left, right = 0, n - 1
-        while left < right:
-            mid = (left + right + 1) >> 1
-            if nums[mid] <= x:
-                left = mid
-            else:
-                right = mid - 1
-        print(f'{t} {left}')
-```
-
-### **Java**
-
-```java
-import java.util.Scanner;
-
-public class Main {
-    public static void main(String[] args) {
-        Scanner sc = new Scanner(System.in);
-        int n = sc.nextInt(), q = sc.nextInt();
-        int[] nums = new int[n];
-        for (int i = 0; i < n; ++i) {
-            nums[i] = sc.nextInt();
-        }
-        while (q-- > 0) {
-            int x = sc.nextInt();
-            int left = 0, right = n - 1;
-            while (left < right) {
-                int mid = (left + right) >> 1;
-                if (nums[mid] >= x) {
-                    right = mid;
-                } else {
-                    left = mid + 1;
-                }
-            }
-            if (nums[left] != x) {
-                System.out.println("-1 -1");
-            } else {
-                int t = left;
-                left = 0;
-                right = n - 1;
-                while (left < right) {
-                    int mid = (left + right + 1) >> 1;
-                    if (nums[mid] <= x) {
-                        left = mid;
-                    } else {
-                        right = mid - 1;
-                    }
-                }
-                System.out.printf("%d %d\n", t, left);
-            }
-        }
-    }
-}
-```
-
-<!-- tabs:end -->
+- [在排序数组中查找元素的第一个和最后一个位置](/solution/0000-0099/0034.Find%20First%20and%20Last%20Position%20of%20Element%20in%20Sorted%20Array/README.md)
diff --git a/basic/searching/BinarySearch/README_EN.md b/basic/searching/BinarySearch/README_EN.md
@@ -0,0 +1,32 @@
+# Binary Search
+
+**Algorithm Templates:**
+
+```java
+/** check x if valid */
+boolean check(int x) {}
+
+/** template1 */
+int binarySearch1(int left, int right) {
+    while (left < right) {
+        int mid = (left + right) >> 1;
+        if (check(mid)) right = mid;
+        else left = mid + 1;
+    }
+    return left;
+}
+
+/** template2 */
+int binarySearch2(int left, int right) {
+    while (left < right) {
+        int mid = (left + right + 1) >> 1;
+        if (check(mid)) left = mid;
+        else right = mid - 1;
+    }
+    return left;
+}
+```
+
+## Examples
+
+- [Find First and Last Position of Element in Sorted Array](/solution/0000-0099/0034.Find%20First%20and%20Last%20Position%20of%20Element%20in%20Sorted%20Array/README_EN.md)