feat: #240 添加python Code

Author: azl397985856

problems/240.search-a-2-d-matrix-ii.md

@@ -43,6 +43,11 @@ Given target = 20, return false.
 
 ## 代码
 
+代码支持：JavaScript, Python3
+
+
+JavaScript Code:
+
 ```js
 
 /*
@@ -52,37 +57,6 @@ Given target = 20, return false.
  *
  * https://leetcode.com/problems/search-a-2d-matrix-ii/description/
  *
- * algorithms
- * Medium (40.30%)
- * Total Accepted:    170K
- * Total Submissions: 419.1K
- * Testcase Example:  '[[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]]\n5'
- *
- * Write an efficient algorithm that searches for a value in an m x n matrix.
- * This matrix has the following properties:
- * 
- * 
- * Integers in each row are sorted in ascending from left to right.
- * Integers in each column are sorted in ascending from top to bottom.
- * 
- * 
- * Example:
- * 
- * Consider the following matrix:
- * 
- * 
- * [
- * ⁠ [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]
- * ]
- * 
- * 
- * Given target = 5, return true.
- * 
- * Given target = 20, return false.
  * 
  */
 /**
@@ -114,3 +88,25 @@ var searchMatrix = function(matrix, target) {
 };
 ```
 
+Python Code:
+
+```python
+class Solution:
+    def searchMatrix(self, matrix, target):
+        m = len(matrix)
+        if m == 0:
+            return False
+        n = len(matrix[0])
+        i = m - 1
+        j = 0
+
+        while i >= 0 and j < n:
+            if matrix[i][j] == target:
+                return True
+            if matrix[i][j] > target:
+                i -= 1
+            else:
+                j += 1
+        return False
+```
+