set matrix zero medium

Author: weiy

Diff for: Array/SetMatrixZeroes.py

@@ -0,0 +1,121 @@
+"""
+Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.
+
+Example 1:
+
+Input: 
+[
+  [1,1,1],
+  [1,0,1],
+  [1,1,1]
+]
+Output: 
+[
+  [1,0,1],
+  [0,0,0],
+  [1,0,1]
+]
+Example 2:
+
+Input: 
+[
+  [0,1,2,0],
+  [3,4,5,2],
+  [1,3,1,5]
+]
+Output: 
+[
+  [0,0,0,0],
+  [0,4,5,0],
+  [0,3,1,0]
+]
+Follow up:
+
+1. A straight forward solution using O(mn) space is probably a bad idea.
+2. A simple improvement uses O(m + n) space, but still not the best solution.
+3. Could you devise a constant space solution?
+
+若某点为 0 则此行与此列均设为0。
+
+关键字：
+1. 坐标去重。
+2. 坐标标记。
+3. 利用原来的空间进行标记。
+
+1. 先说去重，去重是我想到的第一种思路：
+利用set的O(1) 特性去去重，将坐标转换为字符串，然后在转换回去。
+
+这个的效率正如 进阶的 1 所说，O(mn) 空间，bad idea.
+
+能 beat 20% 左右。
+
+2. 坐标标记
+   x     x
+x [0,1,2,0],
+  [3,4,5,2],
+  [1,3,1,5]
+
+只需要标记 x 和 y一整行即可，也正如进阶 2 所说，空间 O(m+n)，good but not best.
+用这个已经可以beat 94%
+
+3. 主要在于四角，稍后在更。
+
+
+测试地址：
+https://leetcode.com/problems/set-matrix-zeroes/description/
+
+"""
+class Solution(object):
+    def setZeroes(self, matrix):
+        """
+        :type matrix: List[List[int]]
+        :rtype: void Do not return anything, modify matrix in-place instead.
+        """
+        # x
+        raw_length = len(matrix[0])
+        # y
+        column_length = len(matrix)
+        
+        # O(mn) 第一版.
+        # {"12", "23", "34"}
+
+#         zero_xy = set()
+        
+#         def makeXY(x, y):
+#             xy = set()
+#             x = str(x)
+#             y = str(y)
+#             for i in range(raw_length):
+#                 xy.add(','.join([str(i), y]))
+            
+#             for i in range(column_length):
+#                 xy.add(','.join([x, str(i)]))
+                
+#             return xy
+        
+#         for y in range(column_length):
+#             for x in range(raw_length):
+#                 if matrix[y][x] == 0:
+#                     zero_xy.update(makeXY(x, y))
+
+#         for i in zero_xy:
+#             t = i.split(',')
+#             matrix[int(t[1])][int(t[0])] = 0
+
+#         第二版 O(M+N)
+#         raw = []
+#         column = []
+        
+#         for y in range(column_length):
+#             for x in range(raw_length):
+#                 if matrix[y][x] == 0:
+#                     raw.append(x)
+#                     column.append(y)
+                    
+#         for x in raw:
+#             for y in range(column_length):
+#                 matrix[y][x] = 0
+        
+#         for y in column:
+#             for x in range(raw_length):
+#                 matrix[y][x] = 0