feat: add python and java solutions to lcof problem

添加《剑指 Offer》题解：面试题56 - II. 数组中数字出现的次数 II

Author: yanglbme

lcof/面试题56 - II. 数组中数字出现的次数 II/README.md

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+# [面试题56 - II. 数组中数字出现的次数 II](https://leetcode-cn.com/problems/shu-zu-zhong-shu-zi-chu-xian-de-ci-shu-ii-lcof/)
+
+## 题目描述
+在一个数组 `nums` 中除一个数字只出现一次之外，其他数字都出现了三次。请找出那个只出现一次的数字。
+
+**示例 1：**
+
+```
+输入：nums = [3,4,3,3]
+输出：4
+```
+
+**示例 2：**
+
+```
+输入：nums = [9,1,7,9,7,9,7]
+输出：1
+```
+
+**限制：**
+
+- `1 <= nums.length <= 10000`
+- `1 <= nums[i] < 2^31`
+
+## 解法
+### Python3
+```python
+class Solution:
+    def singleNumber(self, nums: List[int]) -> int:
+        bits = [0 for _ in range(32)]
+        for num in nums:
+            for i in range(32):
+                bits[i] += (num & 1)
+                num >>= 1
+        
+        res = 0
+        for i in range(32):
+            if bits[i] % 3 == 1:
+                res += pow(2, i)
+        return res
+```
+
+### Java
+```java
+class Solution {
+    public int singleNumber(int[] nums) {
+        int[] bits = new int[32];
+        for (int num : nums) {
+            for (int i = 0; i < 32; ++i) {
+                bits[i] += (num & 1);
+                num >>= 1;
+            }
+        }
+        int res = 0;
+        for (int i = 0; i < 32; ++i) {
+            if (bits[i] % 3 == 1) {
+                res += (int) Math.pow(2, i);
+            }
+        }
+        return res;
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题56 - II. 数组中数字出现的次数 II/Solution.java

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+class Solution {
+    public int singleNumber(int[] nums) {
+        int[] bits = new int[32];
+        for (int num : nums) {
+            for (int i = 0; i < 32; ++i) {
+                bits[i] += (num & 1);
+                num >>= 1;
+            }
+        }
+        int res = 0;
+        for (int i = 0; i < 32; ++i) {
+            if (bits[i] % 3 == 1) {
+                res += (int) Math.pow(2, i);
+            }
+        }
+        return res;
+    }
+}

lcof/面试题56 - II. 数组中数字出现的次数 II/Solution.py

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+class Solution:
+    def singleNumber(self, nums: List[int]) -> int:
+        bits = [0 for _ in range(32)]
+        for num in nums:
+            for i in range(32):
+                bits[i] += (num & 1)
+                num >>= 1
+        
+        res = 0
+        for i in range(32):
+            if bits[i] % 3 == 1:
+                res += pow(2, i)
+        return res