leetcode -easy- 771. Jewels and Stones [Python]

最新推荐文章于 2020-09-06 09:19:39 发布

王后溪

最新推荐文章于 2020-09-06 09:19:39 发布

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本文介绍了一种算法，用于计算给定字符串S中有多少字符同时出现在另一字符串J中，这里J代表宝石类型，S代表拥有的石头。算法通过遍历S，检查每个元素是否在J中，若在则计数加一，最终返回计数值。

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You’re given strings J representing the types of stones that are jewels, and S representing the stones you have. Each character in S is a type of stone you have. You want to know how many of the stones you have are also jewels.
The letters in J are guaranteed distinct, and all characters in J and S are letters. Letters are case sensitive, so “a” is considered a different type of stone from “A”.

思路：遍历S，如果元素也在J中，计数加一：

class Solution(object):
    def numJewelsInStones(self, J, S):
        """
        :type J: str
        :type S: str
        :rtype: int
        """
        count = 0
        for i in S:
            if i in J:
                count = count + 1
        return count

Runtime: 12 ms, faster than 94.14% of Python online submissions for Jewels and Stones.
Memory Usage: 11.6 MB, less than 94.62% of Python online submissions for Jewels and Stones.