feat: add python and java solutions to lcof problem

添加《剑指 Offer》题解：面试题29. 顺时针打印矩阵

Author: yanglbme

lcof/面试题29. 顺时针打印矩阵/README.md

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+# [面试题29. 顺时针打印矩阵](https://leetcode-cn.com/problems/shun-shi-zhen-da-yin-ju-zhen-lcof/)
+
+## 题目描述
+输入一个矩阵，按照从外向里以顺时针的顺序依次打印出每一个数字。
+
+**示例 1：**
+
+```
+输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
+输出：[1,2,3,6,9,8,7,4,5]
+```
+
+**示例 2：**
+
+```
+输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
+输出：[1,2,3,4,8,12,11,10,9,5,6,7]
+```
+
+**限制：**
+
+- `0 <= matrix.length <= 100`
+- `0 <= matrix[i].length <= 100`
+
+## 解法
+### Python3
+```python
+class Solution:
+    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
+        if len(matrix) == 0:
+            return []
+
+        s1, e1, s2, e2 = 0, 0, len(matrix) - 1, len(matrix[0]) - 1
+        res = []
+        while s1 <= s2 and e1 <= e2:
+            res += self._spiral_add(matrix, s1, e1, s2, e2)
+            s1, e1, s2, e2 = s1 + 1, e1 + 1, s2 - 1, e2 - 1
+        return res
+
+    def _spiral_add(self, matrix, s1, e1, s2, e2) -> List[int]:
+        if s1 == s2:
+            return [matrix[s1][j] for j in range(e1, e2 + 1)]
+        if e1 == e2:
+            return [matrix[i][e1] for i in range(s1, s2 + 1)]
+        return [matrix[s1][j] for j in range(e1, e2)] + \
+               [matrix[i][e2] for i in range(s1, s2)] + \
+               [matrix[s2][j] for j in range(e2, e1, -1)] + \
+               [ matrix[i][e1] for i in range(s2, s1, -1)]
+
+```
+
+### Java
+```java
+class Solution {
+    private int index;
+    public int[] spiralOrder(int[][] matrix) {
+        if (matrix.length == 0) {
+            return new int[0];
+        }
+        index = 0;
+        int m = matrix.length, n = matrix[0].length;
+        int s1 = 0, e1 = 0, s2 = m - 1, e2 = n - 1;
+        int[] res = new int[m * n];
+        while (s1 <= s2 && e1 <= e2) {
+            spiralAdd(matrix, s1++, e1++, s2--, e2--, res);
+        }
+        return res;
+    }
+
+    public void spiralAdd(int[][] matrix, int s1, int e1, int s2, int e2, int[] res) {
+        if (s1 == s2) {
+            for (int j = e1; j <= e2; ++j) {
+                res[index++] = matrix[s1][j];
+            }
+            return;
+        }
+        if (e1 == e2) {
+            for (int i = s1; i <= s2; ++i) {
+                res[index++] = matrix[i][e1];
+            }
+            return;
+        }
+
+        for (int j = e1; j < e2; ++j) {
+            res[index++] = matrix[s1][j];
+        }
+        for (int i = s1; i < s2; ++i) {
+            res[index++] = matrix[i][e2];
+        }
+        for (int j = e2; j > e1; --j) {
+            res[index++] = matrix[s2][j];
+        }
+        for (int i = s2; i > s1; --i) {
+            res[index++] = matrix[i][e1];
+        }
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题29. 顺时针打印矩阵/Solution.java

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+class Solution {
+    private int index;
+
+    public int[] spiralOrder(int[][] matrix) {
+        if (matrix.length == 0) {
+            return new int[0];
+        }
+        index = 0;
+        int m = matrix.length, n = matrix[0].length;
+        int s1 = 0, e1 = 0, s2 = m - 1, e2 = n - 1;
+        int[] res = new int[m * n];
+        while (s1 <= s2 && e1 <= e2) {
+            spiralAdd(matrix, s1++, e1++, s2--, e2--, res);
+        }
+        return res;
+    }
+
+    public void spiralAdd(int[][] matrix, int s1, int e1, int s2, int e2, int[] res) {
+        if (s1 == s2) {
+            for (int j = e1; j <= e2; ++j) {
+                res[index++] = matrix[s1][j];
+            }
+            return;
+        }
+        if (e1 == e2) {
+            for (int i = s1; i <= s2; ++i) {
+                res[index++] = matrix[i][e1];
+            }
+            return;
+        }
+
+        for (int j = e1; j < e2; ++j) {
+            res[index++] = matrix[s1][j];
+        }
+        for (int i = s1; i < s2; ++i) {
+            res[index++] = matrix[i][e2];
+        }
+        for (int j = e2; j > e1; --j) {
+            res[index++] = matrix[s2][j];
+        }
+        for (int i = s2; i > s1; --i) {
+            res[index++] = matrix[i][e1];
+        }
+    }
+}

lcof/面试题29. 顺时针打印矩阵/Solution.py

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+class Solution:
+    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
+        if len(matrix) == 0:
+            return []
+
+        s1, e1, s2, e2 = 0, 0, len(matrix) - 1, len(matrix[0]) - 1
+        res = []
+        while s1 <= s2 and e1 <= e2:
+            res += self._spiral_add(matrix, s1, e1, s2, e2)
+            s1, e1, s2, e2 = s1 + 1, e1 + 1, s2 - 1, e2 - 1
+        return res
+
+    def _spiral_add(self, matrix, s1, e1, s2, e2) -> List[int]:
+        if s1 == s2:
+            return [matrix[s1][j] for j in range(e1, e2 + 1)]
+        if e1 == e2:
+            return [matrix[i][e1] for i in range(s1, s2 + 1)]
+        return [matrix[s1][j] for j in range(e1, e2)] + \
+               [matrix[i][e2] for i in range(s1, s2)] + \
+               [matrix[s2][j] for j in range(e2, e1, -1)] + \
+               [ matrix[i][e1] for i in range(s2, s1, -1)]