feat: add python and java solutions to lcof problem

添加《剑指 Offer》题解：面试题21. 调整数组顺序使奇数位于偶数前面

Author: yanglbme

lcof/面试题21. 调整数组顺序使奇数位于偶数前面/README.md

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+# [面试题21. 调整数组顺序使奇数位于偶数前面](https://leetcode-cn.com/problems/diao-zheng-shu-zu-shun-xu-shi-qi-shu-wei-yu-ou-shu-qian-mian-lcof/)
+
+## 题目描述
+输入一个整数数组，实现一个函数来调整该数组中数字的顺序，使得所有奇数位于数组的前半部分，所有偶数位于数组的后半部分。
+
+**示例：**
+
+```
+输入：nums = [1,2,3,4]
+输出：[1,3,2,4] 
+注：[3,1,2,4] 也是正确的答案之一。
+```
+
+**提示：**
+
+- 1 <= nums.length <= 50000
+- 1 <= nums[i] <= 10000
+
+## 解法
+### Python3
+```python
+class Solution:
+    def exchange(self, nums: List[int]) -> List[int]:
+        res = [0 for _ in range(len(nums))]
+        p, q = 0, len(nums) - 1
+        for e in nums:
+            if (e & 1) == 0:
+                res[q] = e
+                q -= 1
+            else:
+                res[p] = e
+                p += 1
+        return res
+```
+
+### Java
+```java
+class Solution {
+    public int[] exchange(int[] nums) {
+        int len = nums.length;
+        int[] res = new int[len];
+        int p = 0, q = len - 1;
+        for (int e : nums) {
+            if ((e & 1) == 0) {
+                res[q--] = e;
+            } else {
+                res[p++] = e;
+            }
+        }
+        return res;
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题21. 调整数组顺序使奇数位于偶数前面/Solution.java

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+class Solution {
+    public int[] exchange(int[] nums) {
+        int len = nums.length;
+        int[] res = new int[len];
+        int p = 0, q = len - 1;
+        for (int e : nums) {
+            if ((e & 1) == 0) {
+                res[q--] = e;
+            } else {
+                res[p++] = e;
+            }
+        }
+        return res;
+    }
+}

lcof/面试题21. 调整数组顺序使奇数位于偶数前面/Solution.py

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+class Solution:
+    def exchange(self, nums: List[int]) -> List[int]:
+        res = [0 for _ in range(len(nums))]
+        p, q = 0, len(nums) - 1
+        for e in nums:
+            if (e & 1) == 0:
+                res[q] = e
+                q -= 1
+            else:
+                res[p] = e
+                p += 1
+        return res