A pull request from cx (#164)

* The original code did not consider the exit condition when the stations cannot perfectly cover the states_needed.

* The code in 09_dynamic_programming/python/01_longest_common_subsequence.py is completed. And the 02_longest_common_substring.py is added.

08_greedy_algorithms/python/01_set_covering.py

@@ -10,16 +10,23 @@ stations["kfive"] = set(["ca", "az"])
 
 final_stations = set()
 
-while states_needed:
-  best_station = None
-  states_covered = set()
-  for station, states_for_station in stations.items():
-    covered = states_needed & states_for_station
-    if len(covered) > len(states_covered):
-      best_station = station
-      states_covered = covered
+def my_set_covering(states_needed, stations):
+    final_stations = set()
+    #while states_needed is not None: 这个不对，Set()而不是None
+    while states_needed:
+        best_station = None
+        states_covered = set()
+        for station, states_for_station in stations.items():
+            covered = states_needed & states_for_station
+            if len(covered) > len(states_covered) and station not in final_stations:
+                best_station = station
+                states_covered = covered
+        if best_station is not None:
+            states_needed -= states_covered
+            final_stations.add(best_station)
+            stations.pop(best_station)
+        else:
+            return None
+    return final_stations
 
-  states_needed -= states_covered
-  final_stations.add(best_station)
-
-print(final_stations)
+print(my_set_covering(states_needed, stations))

09_dynamic_programming/python/01_longest_common_subsequence.py

@@ -1,6 +1,13 @@
-if word_a[i] == word_b[j]:
-  # The letters match.
-  cell[i][j] = cell[i-1][j-1] + 1
-else:
-  # The letters don't match.
-  cell[i][j] = max(cell[i-1][j], cell[i][j-1])
+dp_table_blue = ["b", "l", "u", "e"]
+dp_table_clues = ["c", "l", "u", "e", "s"]
+dp_table = [[0 for i in range(len(dp_table_blue))] for i in range(len(dp_table_clues))] # (5,4)
+print(dp_table)
+
+for i in range(0, len(dp_table_blue)):
+    for j in range(0, len(dp_table_clues)):
+        if dp_table_clues[j] == dp_table_blue[i]:
+            dp_table[j][i] = dp_table[j-1][i-1] + 1
+        else:
+            dp_table[j][i] = max(dp_table[j-1][i], dp_table[j][i-1])
+
+print(dp_table)

09_dynamic_programming/python/02_longest_common_substring.py

@@ -0,0 +1,11 @@
+dp_table_blue = ["b", "l", "u", "e"]
+dp_table_clues = ["c", "l", "u", "e", "s"]
+dp_table = [[0 for i in range(len(dp_table_blue))] for i in range(len(dp_table_clues))] # (5,4)
+print(dp_table)
+
+for i in range(0, len(dp_table_blue)):
+    for j in range(0, len(dp_table_clues)):
+        if dp_table_clues[j] == dp_table_blue[i]:
+            dp_table[i][j] = dp_table[i-1][i-1] + 1
+
+print(dp_table)