feat: add python and java solutions to lcof problem

添加《剑指 Offer》题解：面试题53 - I. 在排序数组中查找数字 I

Author: yanglbme

lcof/README.md

@@ -59,7 +59,7 @@
 |  [51](https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof)  |  [数组中的逆序对](./%E9%9D%A2%E8%AF%95%E9%A2%9851.%20%E6%95%B0%E7%BB%84%E4%B8%AD%E7%9A%84%E9%80%86%E5%BA%8F%E5%AF%B9)  |  困难  |
 |  [52](https://leetcode-cn.com/problems/liang-ge-lian-biao-de-di-yi-ge-gong-gong-jie-dian-lcof)  |  [两个链表的第一个公共节点](./%E9%9D%A2%E8%AF%95%E9%A2%9852.%20%E4%B8%A4%E4%B8%AA%E9%93%BE%E8%A1%A8%E7%9A%84%E7%AC%AC%E4%B8%80%E4%B8%AA%E5%85%AC%E5%85%B1%E8%8A%82%E7%82%B9)  |  简单  |
 |  [53 - II](https://leetcode-cn.com/problems/que-shi-de-shu-zi-lcof)  |  [缺失的数字](./%E9%9D%A2%E8%AF%95%E9%A2%9853%20-%20II.%20%E7%BC%BA%E5%A4%B1%E7%9A%84%E6%95%B0%E5%AD%97)  |  简单  |
-|  [53 - I](https://leetcode-cn.com/problems/zai-pai-xu-shu-zu-zhong-cha-zhao-shu-zi-lcof)  |  [在排序数组中查找数字](./%E9%9D%A2%E8%AF%95%E9%A2%9853%20-%20I.%20%E5%9C%A8%E6%8E%92%E5%BA%8F%E6%95%B0%E7%BB%84%E4%B8%AD%E6%9F%A5%E6%89%BE%E6%95%B0%E5%AD%97)  |  简单  |
+|  [53 - I](https://leetcode-cn.com/problems/zai-pai-xu-shu-zu-zhong-cha-zhao-shu-zi-lcof)  |  [在排序数组中查找数字](./%E9%9D%A2%E8%AF%95%E9%A2%9853%20-%20I.%20%E5%9C%A8%E6%8E%92%E5%BA%8F%E6%95%B0%E7%BB%84%E4%B8%AD%E6%9F%A5%E6%89%BE%E6%95%B0%E5%AD%97%20I)  |  简单  |
 |  [54](https://leetcode-cn.com/problems/er-cha-sou-suo-shu-de-di-kda-jie-dian-lcof)  |  [二叉搜索树的第k大节点](./%E9%9D%A2%E8%AF%95%E9%A2%9854.%20%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E7%9A%84%E7%AC%ACk%E5%A4%A7%E8%8A%82%E7%82%B9)  |  简单  |
 |  [55 - II](https://leetcode-cn.com/problems/ping-heng-er-cha-shu-lcof)  |  [平衡二叉树](./%E9%9D%A2%E8%AF%95%E9%A2%9855%20-%20II.%20%E5%B9%B3%E8%A1%A1%E4%BA%8C%E5%8F%89%E6%A0%91)  |  简单  |
 |  [55 - I](https://leetcode-cn.com/problems/er-cha-shu-de-shen-du-lcof)  |  [二叉树的深度](./%E9%9D%A2%E8%AF%95%E9%A2%9855%20-%20I.%20%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E6%B7%B1%E5%BA%A6)  |  简单  |

lcof/面试题53 - I. 在排序数组中查找数字 I/README.md

@@ -0,0 +1,104 @@
+# [面试题53 - I. 在排序数组中查找数字 I](https://leetcode-cn.com/problems/zai-pai-xu-shu-zu-zhong-cha-zhao-shu-zi-lcof/)
+
+## 题目描述
+统计一个数字在排序数组中出现的次数。
+
+**示例 1:**
+
+```
+输入: nums = [5,7,7,8,8,10], target = 8
+输出: 2
+```
+
+**示例 2:**
+
+```
+输入: nums = [5,7,7,8,8,10], target = 6
+输出: 0
+```
+
+**限制：**
+
+- `0 <= 数组长度 <= 50000`
+
+## 解法
+### Python3
+```python
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        if not nums:
+            return 0
+        l, r = 0, len(nums) - 1
+        while l <= r:
+            m = l + ((r - l) >> 1)
+            if nums[m] == target:
+                return self._count(nums, l, r, m)
+            if nums[m] < target:
+                l = m + 1
+            else:
+                r = m - 1
+        return 0
+    
+    def _count(self, nums, l, r, m) -> int:
+        cnt = 0
+        for i in range(m, l - 1, -1):
+            if nums[i] == nums[m]:
+                cnt += 1
+            elif nums[i] < nums[m]:
+                break
+        
+        for i in range(m + 1, r + 1):
+            if nums[i] == nums[m]:
+                cnt += 1
+            elif nums[i] > nums[m]:
+                break
+        return cnt
+```
+
+### Java
+```java
+class Solution {
+    public int search(int[] nums, int target) {
+        if (nums.length == 0) {
+            return 0;
+        }
+        int l = 0, r = nums.length - 1;
+        while (l <= r) {
+            int m = l + ((r - l) >> 1);
+            if (nums[m] == target) {
+                return count(nums, l, r, m);
+            }
+            if (nums[m] < target) {
+                l = m + 1;
+            } else {
+                r = m - 1;
+            }
+        }
+        return 0;
+    }
+
+    private int count(int[] nums, int l, int r, int m) {
+        int cnt = 0;
+        for (int i = m; i >= l; --i) {
+            if (nums[i] == nums[m]) {
+                ++cnt;
+            } else if (nums[i] < nums[m]) {
+                break;
+            }
+        }
+        for (int i = m + 1; i <= r; ++i) {
+            if (nums[i] == nums[m]) {
+                ++cnt;
+            } else if (nums[i] > nums[m]) {
+                break;
+            }
+        }
+        return cnt;
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题53 - I. 在排序数组中查找数字 I/Solution.java

@@ -0,0 +1,39 @@
+class Solution {
+    public int search(int[] nums, int target) {
+        if (nums.length == 0) {
+            return 0;
+        }
+        int l = 0, r = nums.length - 1;
+        while (l <= r) {
+            int m = l + ((r - l) >> 1);
+            if (nums[m] == target) {
+                return count(nums, l, r, m);
+            }
+            if (nums[m] < target) {
+                l = m + 1;
+            } else {
+                r = m - 1;
+            }
+        }
+        return 0;
+    }
+
+    private int count(int[] nums, int l, int r, int m) {
+        int cnt = 0;
+        for (int i = m; i >= l; --i) {
+            if (nums[i] == nums[m]) {
+                ++cnt;
+            } else if (nums[i] < nums[m]) {
+                break;
+            }
+        }
+        for (int i = m + 1; i <= r; ++i) {
+            if (nums[i] == nums[m]) {
+                ++cnt;
+            } else if (nums[i] > nums[m]) {
+                break;
+            }
+        }
+        return cnt;
+    }
+}

lcof/面试题53 - I. 在排序数组中查找数字 I/Solution.py

@@ -0,0 +1,29 @@
+class Solution:
+    def search(self, nums: List[int], target: int) -> int:
+        if not nums:
+            return 0
+        l, r = 0, len(nums) - 1
+        while l <= r:
+            m = l + ((r - l) >> 1)
+            if nums[m] == target:
+                return self._count(nums, l, r, m)
+            if nums[m] < target:
+                l = m + 1
+            else:
+                r = m - 1
+        return 0
+    
+    def _count(self, nums, l, r, m) -> int:
+        cnt = 0
+        for i in range(m, l - 1, -1):
+            if nums[i] == nums[m]:
+                cnt += 1
+            elif nums[i] < nums[m]:
+                break
+        
+        for i in range(m + 1, r + 1):
+            if nums[i] == nums[m]:
+                cnt += 1
+            elif nums[i] > nums[m]:
+                break
+        return cnt