771. Jewels and Stones

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发布于 2019-05-26 00:29:55

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发布于 2019-05-26 00:29:55

文章被收录于专栏：机器学习入门

LWC 69: 771. Jewels and Stones

Problem:

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.

思路：记录J的信息，遍历S，采用Hashmap的思路，每次查询O(1)，总时间复杂度为O(n)

Java版本：

    public int numJewelsInStones(String J, String S) {
        int[] count = new int[64];
        for (char c : J.toCharArray()) {
            count[c - 'A']++;
        }
        int ans = 0;
        for (char c : S.toCharArray()) {
            if (count[c - 'A'] >= 1) ans ++;
        }
        return ans;
    }

Python版本：

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

再精简一波：

class Solution(object):
    def numJewelsInStones(self, J, S):
        """
        :type J: str
        :type S: str
        :rtype: int
        """
        info = set(J)
        return sum(c in info for c in S)

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原始发表：2018年01月29日，如有侵权请联系 cloudcommunity@tencent.com 删除

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