feat: add python and java solutions to lcof problem

添加《剑指 Offer》题解：面试题39. 数组中出现次数超过一半的数字

Author: yanglbme

lcof/面试题39. 数组中出现次数超过一半的数字/README.md

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+# [面试题39. 数组中出现次数超过一半的数字](https://leetcode-cn.com/problems/shu-zu-zhong-chu-xian-ci-shu-chao-guo-yi-ban-de-shu-zi-lcof/)
+
+## 题目描述
+数组中有一个数字出现的次数超过数组长度的一半，请找出这个数字。
+
+你可以假设数组是非空的，并且给定的数组总是存在多数元素。
+
+**示例 1:**
+
+```
+输入: [1, 2, 3, 2, 2, 2, 5, 4, 2]
+输出: 2
+```
+
+**限制：**
+
+- `1 <= 数组长度 <= 50000`
+
+## 解法
+摩尔投票法。时间复杂度 O(n)，空间复杂度 O(1)。
+
+### Python3
+```python
+class Solution:
+    def majorityElement(self, nums: List[int]) -> int:
+        cnt = major = 0
+        for num in nums:
+            if cnt == 0:
+                major = num
+                cnt += 1
+            else:
+                cnt += (1 if major == num else -1)
+        return major
+```
+
+### Java
+```java
+class Solution {
+    public int majorityElement(int[] nums) {
+        int major = 0, cnt = 0;
+        for (int num : nums) {
+            if (cnt == 0) {
+                major = num;
+                ++cnt;
+            } else {
+                cnt += (num == major ? 1 : -1);
+            }
+        }
+        return major;
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题39. 数组中出现次数超过一半的数字/Solution.java

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+class Solution {
+    public int majorityElement(int[] nums) {
+        int major = 0, cnt = 0;
+        for (int num : nums) {
+            if (cnt == 0) {
+                major = num;
+                ++cnt;
+            } else {
+                cnt += (num == major ? 1 : -1);
+            }
+        }
+        return major;
+    }
+}

lcof/面试题39. 数组中出现次数超过一半的数字/Solution.py

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+class Solution:
+    def majorityElement(self, nums: List[int]) -> int:
+        cnt = major = 0
+        for num in nums:
+            if cnt == 0:
+                major = num
+                cnt += 1
+            else:
+                cnt += (1 if major == num else -1)
+        return major