LeetCode:771. Jewels and Stones

最新推荐文章于 2020-12-27 23:49:46 发布

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本文介绍了解决LeetCode上编号为771的问题“宝石与石头”的两种方法。首先，通过暴力求解的方式，使用双重循环或in关键词进行计数；其次，提出了一种更高效的哈希方法，利用ASCII码特性创建哈希表加速查找过程。

LeetCode:771. Jewels and Stones

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

其实就是统计J中的字符在S中的出现的次数。

思路一：暴力求解

双重循环，不再赘述。Python 代码实现

或者内部循环里直接使用 in 关键词：

class Solution:
    def numJewelsInStones(self, J: str, S: str) -> int:
        lS = len(S)
        countJ = 0
        for i in range(lS):
            if (S[i] in J):
                countJ+=1
        return countJ

思路二：使用ASCII码的哈希方法

小写字符 a 的ASCII码为97，大小写字母一共58个。所以要用到一个长度为58的整型数组，如果J字符串中还有某个字符，则对应的位置设为1。然后遍历一次字符串S，如果某个位置的字符对应的哈希值为1，则表明存在于J中，个数加1。

Python 代码实现

class Solution:
    def numJewelsInStones(self, J: str, S: str) -> int:
        lJ = len(J)
        lS = len(S)
        charMap = [0]*58
        for i in range(lJ):
            charMap[ord(J[i])-97]=1
        countJ = 0
        for i in range(lS):
            if (charMap[ord(S[i])-97] == 1):
                countJ+=1
        return countJ