feat: add python and java solutions to lcof problem

添加《剑指 Offer》题解：面试题53 - II. 0～n-1中缺失的数字

Author: yanglbme

README_CN.md

@@ -46,4 +46,4 @@
 <a href="https://opencollective.com/doocs-leetcode/contributors.svg?width=890&button=true"><img src="https://opencollective.com/doocs-leetcode/contributors.svg?width=890&button=false" /></a>
 
 ## 许可证
-<a rel="license" href="https://github.com/doocs/leetcode/blob/master/LICENSE"><img alt="Creative Commons License" style="border-width:0" src="./img/cc-by-sa-88x31.png" /></a><br /><a rel="license" href="http://creativecommons.org/licenses/by-sa/4.0/">知识共享 版权归属-相同方式共享 4.0 国际 公共许可证</a>。
+<a rel="license" href="https://github.com/doocs/leetcode/blob/master/LICENSE"><img alt="Creative Commons License" style="border-width:0" src="./img/cc-by-sa-88x31.png" /></a><br /><a rel="license" href="http://creativecommons.org/licenses/by-sa/4.0/">知识共享 版权归属-相同方式共享 4.0 国际 公共许可证</a>

lcof/README.md

@@ -58,7 +58,7 @@
 |  [50](https://leetcode-cn.com/problems/di-yi-ge-zhi-chu-xian-yi-ci-de-zi-fu-lcof)  |  [第一个只出现一次的字符](./%E9%9D%A2%E8%AF%95%E9%A2%9850.%20%E7%AC%AC%E4%B8%80%E4%B8%AA%E5%8F%AA%E5%87%BA%E7%8E%B0%E4%B8%80%E6%AC%A1%E7%9A%84%E5%AD%97%E7%AC%A6)  |  简单  |
 |  [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 - 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.%200%EF%BD%9En-1%E4%B8%AD%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%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  |

lcof/面试题53 - II. 0～n-1中缺失的数字/README.md

@@ -0,0 +1,70 @@
+# [面试题53 - II. 0～n-1中缺失的数字](https://leetcode-cn.com/problems/que-shi-de-shu-zi-lcof/)
+
+## 题目描述
+一个长度为n-1的递增排序数组中的所有数字都是唯一的，并且每个数字都在范围0～n-1之内。在范围0～n-1内的n个数字中有且只有一个数字不在该数组中，请找出这个数字。
+
+**示例 1:**
+
+```
+输入: [0,1,3]
+输出: 2
+```
+
+**示例 2:**
+
+```
+输入: [0,1,2,3,4,5,6,7,9]
+输出: 8
+```
+
+**限制：**
+
+- `1 <= 数组长度 <= 10000`
+
+## 解法
+### Python3
+```python
+class Solution:
+    def missingNumber(self, nums: List[int]) -> int:
+        l, r = 0, len(nums) - 1
+        if r == 0 or nums[0] == 1:
+            return nums[0] ^ 1
+        if nums[r] == r:
+            return r + 1
+        while r - l > 1:
+            m = l + ((r - l) >> 1)
+            if nums[m] == m:
+                l = m
+            else:
+                r = m
+        return nums[r] - 1
+```
+
+### Java
+```java
+class Solution {
+    public int missingNumber(int[] nums) {
+        int l = 0, r = nums.length - 1;
+        if (r == 0 || nums[0] == 1) {
+            return nums[0] ^ 1;
+        }
+        if (nums[r] == r) {
+            return r + 1;
+        }
+        while (r - l > 1) {
+            int m = l + ((r - l) >> 1);
+            if (nums[m] == m) {
+                l = m;
+            } else {
+                r = m;
+            }
+        }
+        return nums[r] - 1;
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题53 - II. 0～n-1中缺失的数字/Solution.java

@@ -0,0 +1,20 @@
+class Solution {
+    public int missingNumber(int[] nums) {
+        int l = 0, r = nums.length - 1;
+        if (r == 0 || nums[0] == 1) {
+            return nums[0] ^ 1;
+        }
+        if (nums[r] == r) {
+            return r + 1;
+        }
+        while (r - l > 1) {
+            int m = l + ((r - l) >> 1);
+            if (nums[m] == m) {
+                l = m;
+            } else {
+                r = m;
+            }
+        }
+        return nums[r] - 1;
+    }
+}

lcof/面试题53 - II. 0～n-1中缺失的数字/Solution.py

@@ -0,0 +1,14 @@
+class Solution:
+    def missingNumber(self, nums: List[int]) -> int:
+        l, r = 0, len(nums) - 1
+        if r == 0 or nums[0] == 1:
+            return nums[0] ^ 1
+        if nums[r] == r:
+            return r + 1
+        while r - l > 1:
+            m = l + ((r - l) >> 1)
+            if nums[m] == m:
+                l = m
+            else:
+                r = m
+        return nums[r] - 1