LeetCode：771. Jewels and Stones

最新推荐文章于 2019-06-08 15:01:59 发布

原创最新推荐文章于 2019-06-08 15:01:59 发布 · 402 阅读

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本文介绍了一个简单的编程问题：计算给定字符串中属于“宝石”的字符数量。提供了两种解决方案，一种是通过双重循环逐个比较字符，另一种是使用Python内置函数进行高效计算。

题目：
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”.

Example 1:

Input: J = “aA”, S = “aAAbbbb”
Output: 3
Example 2:

Input: J = “z”, S = “ZZ”
Output: 0
Note:

S and J will consist of letters and have length at most 50.
The characters in J are distinct.

Answer：
我的答案：

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

其它答案用到map()和string.count()：

    def numJewelsInStones(self, J, S):
        return sum(map(J.count, S))

    def numJewelsInStones(self, J, S):
        return sum(map(S.count, J))  # this one after seeing https://discuss.leetcode.com/post/244105

    def numJewelsInStones(self, J, S):
        return sum(s in J for s in S)