批改

homework_01.py

+
+# 时间复杂度较高 O(N^2)
 def numJewelsInStones(J, S):
     # count = 0
     # for i in S:
@@ -8,3 +10,29 @@ def numJewelsInStones(J, S):
 
 
 print(numJewelsInStones("aA", "aAAbbbb"))  # 预期输出 3
+
+
+
+
+# o(n)的版本
+def numJewelsInStones(J, S):
+    stores = {}
+    for c in S:
+        if c in stores:
+            stores[c] = stores[c] + 1
+        else:
+            stores[c] = 1
+
+    count = 0
+    for c in J:
+        if c in stores:
+            count += stores[c]
+    print(count)
+
+
+J = "aA"
+S = "aAAbbbb"
+numJewelsInStones(J, S)
+J = "z"
+S = "ZZ"
+numJewelsInStones(J, S)