feat: add python and java solutions to lcof problem

添加《剑指 Offer》题解：面试题57. 和为s的两个数字

Author: yanglbme

lcof/面试题57. 和为s的两个数字/README.md

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+# [面试题57. 和为s的两个数字](https://leetcode-cn.com/problems/he-wei-sde-liang-ge-shu-zi-lcof/)
+
+## 题目描述
+输入一个递增排序的数组和一个数字 s，在数组中查找两个数，使得它们的和正好是 s。如果有多对数字的和等于 s，则输出任意一对即可。
+
+**示例 1：**
+
+```
+输入：nums = [2,7,11,15], target = 9
+输出：[2,7] 或者 [7,2]
+```
+
+**示例 2：**
+
+```
+输入：nums = [10,26,30,31,47,60], target = 40
+输出：[10,30] 或者 [30,10]
+```
+
+**限制：**
+
+- `1 <= nums.length <= 10^5`
+- `1 <= nums[i] <= 10^6`
+
+## 解法
+### Python3
+```python
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        p, q = 0, len(nums) - 1
+        while p < q:
+            s = nums[p] + nums[q]
+            if s == target:
+                return [nums[p], nums[q]]
+            if s < target: 
+                p += 1
+            else: 
+                q -= 1
+
+```
+
+### Java
+```java
+class Solution {
+    public int[] twoSum(int[] nums, int target) {
+        int p = 0, q = nums.length - 1;
+        while (p < q) {
+            int s = nums[p] + nums[q];
+            if (s == target) {
+                return new int[] {nums[p], nums[q]};
+            }
+            if (s < target) {
+                ++p;
+            } else {
+                --q;
+            }
+        }
+        return new int[]{0};
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题57. 和为s的两个数字/Solution.java

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+class Solution {
+    public int[] twoSum(int[] nums, int target) {
+        int p = 0, q = nums.length - 1;
+        while (p < q) {
+            int s = nums[p] + nums[q];
+            if (s == target) {
+                return new int[] {nums[p], nums[q]};
+            }
+            if (s < target) {
+                ++p;
+            } else {
+                --q;
+            }
+        }
+        return new int[]{0};
+    }
+}

lcof/面试题57. 和为s的两个数字/Solution.py

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+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        p, q = 0, len(nums) - 1
+        while p < q:
+            s = nums[p] + nums[q]
+            if s == target:
+                return [nums[p], nums[q]]
+            if s < target: 
+                p += 1
+            else: 
+                q -= 1