commit homework for 宝石与石头

1_homework.py

+# 解题思路，循环拿J的元素与S中的元素对比，如果相等则+1，自己最开始的解法，中规中矩^_^
+def numJewelsInStones(J, S):
+    i = 0
+    l = len(S)
+    for x in J:
+        for xx in range(l):
+            #print(S[xx])
+            if (x == S[xx]):
+               i+=1
+        else:
+            continue
+    print(i)
+
+
+numJewelsInStones("aA","aAAbbbb")
+numJewelsInStones("z","ZZ")
+numJewelsInStones("zqwe","aAbBCcddSSDDGTwWeEeee")
+
+
+# 解题思路，与上面不同，由于J中元素唯一，则反过来用S中的元素来判断是否存在于J中，存在则+1(借鉴了网友的思路，学习了)
+def numJewelsInStones1(J, S):
+    i = 0
+    for s in S:
+        if s in J:
+            i += 1
+    print(i)
+
+numJewelsInStones1("aA","aAAbbbb")
+numJewelsInStones1("z","ZZ")
+numJewelsInStones1("zqwe","aAbBCcddSSDDGTwWeEeee")
+
+# 将字符转为了列表，统计同时存在J和S中元素的个数(借鉴了网友的思路，学习了)
+def numJewelsInStones2(J, S):
+     print(sum([s in J for s in S]))
+
+numJewelsInStones2("aA","aAAbbbb")
+numJewelsInStones2("z","ZZ")
+numJewelsInStones2("zqwe","aAbBCcddSSDDGTwWeEeee")
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