描述
You're given strings jewels representing the types of stones that are jewels, and stones representing the stones you have. Each character in stones is a type of stone you have. You want to know how many of the stones you have are also jewels.
Letters are case sensitive, so "a" is considered a different type of stone from "A".
Example 1:
Input: jewels = "aA", stones = "aAAbbbb"
Output: 3
Example 2:
Input: jewels = "z", stones = "ZZ"
Output: 0
Note:
- 1 <= jewels.length, stones.length <= 50
- jewels and stones consist of only English letters.
- All the characters of jewels are unique.
解析
根据题意，只需要使用字典结构 c 统计 stones 中的字符及其出现个数，然后遍历 jewels ，如果其中的字符在 c 中出现，则取出对应的出现个数与 res 相加，遍历结束得到的最后的结果就是答案。
解答
class Solution(object):
def numJewelsInStones(self, jewels, stones):
"""
:type jewels: str
:type stones: str
:rtype: int
"""
res = 0
c = collections.Counter(stones)
for j in jewels:
if j in c:
res += c.get(j)
return res
运行结果
Runtime: 12 ms, faster than 96.47% of Python online submissions for Jewels and Stones.
Memory Usage: 13.6 MB, less than 8.33% of Python online submissions for Jewels and Stones.
原题链接：leetcode.com/problems/je…
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