feat: add python and java solution to lcof problem

添加《剑指 Offer》题解：面试题12. 矩阵中的路径

Author: yanglbme

lcof/面试题12. 矩阵中的路径/README.md

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+# [面试题12. 矩阵中的路径](https://leetcode-cn.com/problems/ju-zhen-zhong-de-lu-jing-lcof/)
+
+## 题目描述
+请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一格开始，每一步可以在矩阵中向左、右、上、下移动一格。如果一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子。例如，在下面的3×4的矩阵中包含一条字符串“bfce”的路径（路径中的字母用加粗标出）。
+
+```
+[["a","b","c","e"],
+["s","f","c","s"],
+["a","d","e","e"]]
+```
+
+但矩阵中不包含字符串“abfb”的路径，因为字符串的第一个字符 b 占据了矩阵中的第一行第二个格子之后，路径不能再次进入这个格子。
+
+
+**示例 1：**
+
+```
+输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
+输出：true
+```
+
+**示例 2：**
+
+```
+输入：board = [["a","b"],["c","d"]], word = "abcd"
+输出：false
+```
+
+**提示：**
+
+- `1 <= board.length <= 200`
+- `1 <= board[i].length <= 200`
+
+## 解法
+### Python3
+```python
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        if not word:
+            return False
+        rows, cols = len(board), len(board[0])
+        visited = [[False for _ in range(cols)] for _ in range(rows)]
+        for i in range(rows):
+            for j in range(cols):
+                if board[i][j] != word[0]:
+                    continue
+                if self.visit(board, visited, i, j, rows, cols, word):
+                    return True
+        return False
+    
+    def visit(self, board, visited, i, j, rows, cols, word) -> bool:
+        if not word:
+            return True
+        if i < 0 or j < 0 or i >= rows or j >= cols or visited[i][j] or board[i][j] != word[0]:
+            return False
+        visited[i][j] = True
+        res = self.visit(board, visited, i - 1, j, rows, cols, word[1:]) or self.visit(board, visited, i + 1, j, rows, cols, word[1:]) or self.visit(board, visited, i, j - 1, rows, cols, word[1:]) or self.visit(board, visited, i, j + 1, rows, cols, word[1:])
+        visited[i][j] = res
+        return res
+```
+
+### Java
+```java
+class Solution {
+    public boolean exist(char[][] board, String word) {
+        if (word == null || word.length() == 0) {
+            return false;
+        }
+        int rows = board.length, cols = board[0].length;
+        boolean[][] visited = new boolean[rows][cols];
+        for (int i = 0; i < rows; ++i) {
+            for (int j = 0; j < cols; ++j) {
+                if (board[i][j] != word.charAt(0)) {
+                    continue;
+                }
+                if (visit(board, visited, i, j, rows, cols, word)) {
+                    return true;
+                }
+            }
+        }
+        return false;
+    }
+
+    private boolean visit(char[][] board, boolean[][] visited, int i, int j, int rows, int cols, String word) {
+        if (word.length() == 0) {
+            return true;
+        }
+        if (i < 0 || j < 0 || i >= rows || j >= cols || visited[i][j] || board[i][j] != word.charAt(0)) {
+            return false;
+        }
+
+        visited[i][j] = true;
+        String sub = word.substring(1);
+        boolean res = visit(board, visited, i + 1, j, rows, cols, sub)
+                || visit(board, visited, i - 1, j, rows, cols, sub)
+                || visit(board, visited, i, j + 1, rows, cols, sub)
+                || visit(board, visited, i, j - 1, rows, cols, sub);
+        visited[i][j] = res;
+        return res;
+
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题12. 矩阵中的路径/Solution.java

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+class Solution {
+    public boolean exist(char[][] board, String word) {
+        if (word == null || word.length() == 0) {
+            return false;
+        }
+        int rows = board.length, cols = board[0].length;
+        boolean[][] visited = new boolean[rows][cols];
+        for (int i = 0; i < rows; ++i) {
+            for (int j = 0; j < cols; ++j) {
+                if (board[i][j] != word.charAt(0)) {
+                    continue;
+                }
+                if (visit(board, visited, i, j, rows, cols, word)) {
+                    return true;
+                }
+            }
+        }
+        return false;
+    }
+
+    private boolean visit(char[][] board, boolean[][] visited, int i, int j, int rows, int cols, String word) {
+        if (word.length() == 0) {
+            return true;
+        }
+        if (i < 0 || j < 0 || i >= rows || j >= cols || visited[i][j] || board[i][j] != word.charAt(0)) {
+            return false;
+        }
+
+        visited[i][j] = true;
+        String sub = word.substring(1);
+        boolean res = visit(board, visited, i + 1, j, rows, cols, sub)
+                || visit(board, visited, i - 1, j, rows, cols, sub)
+                || visit(board, visited, i, j + 1, rows, cols, sub)
+                || visit(board, visited, i, j - 1, rows, cols, sub);
+        visited[i][j] = res;
+        return res;
+
+    }
+}

lcof/面试题12. 矩阵中的路径/Solution.py

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+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        if not word:
+            return False
+        rows, cols = len(board), len(board[0])
+        visited = [[False for _ in range(cols)] for _ in range(rows)]
+        for i in range(rows):
+            for j in range(cols):
+                if board[i][j] != word[0]:
+                    continue
+                if self.visit(board, visited, i, j, rows, cols, word):
+                    return True
+        return False
+    
+    def visit(self, board, visited, i, j, rows, cols, word) -> bool:
+        if not word:
+            return True
+        if i < 0 or j < 0 or i >= rows or j >= cols or visited[i][j] or board[i][j] != word[0]:
+            return False
+        visited[i][j] = True
+        res = self.visit(board, visited, i - 1, j, rows, cols, word[1:]) or self.visit(board, visited, i + 1, j, rows, cols, word[1:]) or self.visit(board, visited, i, j - 1, rows, cols, word[1:]) or self.visit(board, visited, i, j + 1, rows, cols, word[1:])
+        visited[i][j] = res
+        return res