feat: add python and java solutions to lcof problem

添加《剑指 Offer》题解：面试题56 - I. 数组中数字出现的次数

Author: yanglbme

lcof/面试题56 - I. 数组中数字出现的次数/README.md

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+# [面试题56 - I. 数组中数字出现的次数](https://leetcode-cn.com/problems/shu-zu-zhong-shu-zi-chu-xian-de-ci-shu-lcof/)
+
+## 题目描述
+一个整型数组 nums 里除两个数字之外，其他数字都出现了两次。请写程序找出这两个只出现一次的数字。要求时间复杂度是O(n)，空间复杂度是O(1)。
+
+**示例 1：**
+
+```
+输入：nums = [4,1,4,6]
+输出：[1,6] 或 [6,1]
+```
+
+**示例 2：**
+
+```
+输入：nums = [1,2,10,4,1,4,3,3]
+输出：[2,10] 或 [10,2]
+```
+
+**限制：**
+
+- `2 <= nums <= 10000`
+
+## 解法
+### Python3
+```python
+class Solution:
+    def singleNumbers(self, nums: List[int]) -> List[int]:
+        xor_res = 0
+        for num in nums:
+            xor_res ^= num
+        pos = 0
+        while (xor_res & 1) == 0:
+            pos += 1
+            xor_res >>= 1
+        
+        a = b = 0
+        for num in nums:
+            t = num >> pos
+            if (t & 1) == 0:
+                a ^= num
+            else:
+                b ^= num
+        return [a, b]
+```
+
+### Java
+```java
+class Solution {
+    public int[] singleNumbers(int[] nums) {
+        int xor = 0;
+        for (int num : nums) {
+            xor ^= num;
+        }
+        int pos = 0;
+        while ((xor & 1) == 0) {
+            ++pos;
+            xor >>= 1;
+        }
+        int a = 0, b = 0;
+        for (int num : nums) {
+            int t = num >> pos;
+            if ((t & 1) == 0) {
+                a ^= num;
+            } else {
+                b ^= num;
+            }
+        }
+        return new int[] {a, b};
+    }
+}
+```
+
+### ...
+```
+
+```

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

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+class Solution {
+    public int[] singleNumbers(int[] nums) {
+        int xor = 0;
+        for (int num : nums) {
+            xor ^= num;
+        }
+        int pos = 0;
+        while ((xor & 1) == 0) {
+            ++pos;
+            xor >>= 1;
+        }
+        int a = 0, b = 0;
+        for (int num : nums) {
+            int t = num >> pos;
+            if ((t & 1) == 0) {
+                a ^= num;
+            } else {
+                b ^= num;
+            }
+        }
+        return new int[] { a, b };
+    }
+}

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

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+class Solution:
+    def singleNumbers(self, nums: List[int]) -> List[int]:
+        xor_res = 0
+        for num in nums:
+            xor_res ^= num
+        pos = 0
+        while (xor_res & 1) == 0:
+            pos += 1
+            xor_res >>= 1
+        
+        a = b = 0
+        for num in nums:
+            t = num >> pos
+            if (t & 1) == 0:
+                a ^= num
+            else:
+                b ^= num
+        return [a, b]