feat: add two solutions for lcof problems

添加两道《剑指 Offer》题解

Author: yanglbme

lcof/面试题04. 二维数组中的查找/README.md

@@ -0,0 +1,78 @@
+# [面试题04. 二维数组中的查找](https://leetcode-cn.com/problems/er-wei-shu-zu-zhong-de-cha-zhao-lcof/)
+
+## 题目描述
+在一个 n * m 的二维数组中，每一行都按照从左到右递增的顺序排序，每一列都按照从上到下递增的顺序排序。请完成一个函数，输入这样的一个二维数组和一个整数，判断数组中是否含有该整数。
+
+**示例:**
+
+现有矩阵 matrix 如下：
+
+```
+[
+  [1,   4,  7, 11, 15],
+  [2,   5,  8, 12, 19],
+  [3,   6,  9, 16, 22],
+  [10, 13, 14, 17, 24],
+  [18, 21, 23, 26, 30]
+]
+```
+
+给定 target = 5，返回 `true`。
+
+给定 target = 20，返回 `false`。
+
+**限制：**
+
+- `0 <= n <= 1000`
+
+- `0 <= m <= 1000`
+
+## 解法
+### Python3
+```python
+class Solution:
+    def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
+        if not matrix:
+            return False
+        
+        i = len(matrix) - 1
+        j = 0
+        while i >= 0 and j < len(matrix[0]):
+            if matrix[i][j] == target:
+                return True
+            if matrix[i][j] < target:
+                j += 1
+            else:
+                i -= 1
+        return False
+```
+
+### Java
+```java
+class Solution {
+    public boolean findNumberIn2DArray(int[][] matrix, int target) {
+        if (matrix == null || matrix.length == 0) {
+            return false;
+        }
+        int rows = matrix.length, cols = matrix[0].length;
+
+        int i = rows - 1, j = 0;
+        while (i >= 0 && j < cols) {
+            if (matrix[i][j] == target) {
+                return true;
+            }
+            if (matrix[i][j] < target) {
+                ++j;
+            } else {
+                --i;
+            }
+        }
+        return false;
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题04. 二维数组中的查找/Solution.java

@@ -0,0 +1,21 @@
+class Solution {
+    public boolean findNumberIn2DArray(int[][] matrix, int target) {
+        if (matrix == null || matrix.length == 0) {
+            return false;
+        }
+        int rows = matrix.length, cols = matrix[0].length;
+
+        int i = rows - 1, j = 0;
+        while (i >= 0 && j < cols) {
+            if (matrix[i][j] == target) {
+                return true;
+            }
+            if (matrix[i][j] < target) {
+                ++j;
+            } else {
+                --i;
+            }
+        }
+        return false;
+    }
+}

lcof/面试题04. 二维数组中的查找/Solution.py

@@ -0,0 +1,15 @@
+class Solution:
+    def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
+        if not matrix:
+            return False
+        
+        i = len(matrix) - 1
+        j = 0
+        while i >= 0 and j < len(matrix[0]):
+            if matrix[i][j] == target:
+                return True
+            if matrix[i][j] < target:
+                j += 1
+            else:
+                i -= 1
+        return False

lcof/面试题05. 替换空格/README.md

@@ -0,0 +1,37 @@
+# [面试题05. 替换空格](https://leetcode-cn.com/problems/ti-huan-kong-ge-lcof/)
+
+## 题目描述
+请实现一个函数，把字符串 `s` 中的每个空格替换成"%20"。
+
+**示例 1：**
+
+```
+输入：s = "We are happy."
+输出："We%20are%20happy."
+``` 
+
+**限制：**
+
+- `0 <= s 的长度 <= 10000`
+
+## 解法
+### Python3
+```python
+class Solution:
+    def replaceSpace(self, s: str) -> str:
+        return s.replace(' ', '%20')
+```
+
+### Java
+```java
+class Solution {
+    public String replaceSpace(String s) {
+        return s.replaceAll(" ", "%20");
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题05. 替换空格/Solution.java

@@ -0,0 +1,5 @@
+class Solution {
+    public String replaceSpace(String s) {
+        return s.replaceAll(" ", "%20");
+    }
+}

lcof/面试题05. 替换空格/Solution.py

@@ -0,0 +1,3 @@
+class Solution:
+    def replaceSpace(self, s: str) -> str:
+        return s.replace(' ', '%20')