算法设计与分析：Jewels and Stones（Week 1）

最新推荐文章于 2019-06-17 13:55:16 发布

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本文介绍了一种使用C#实现的高效算法，用于计算给定字符串中属于特定类型（珠宝）字符的数量。该算法通过建立一个布尔数组来标记哪些字符是有效的珠宝字符，并通过对输入字符串进行一次遍历即可得出结果。

学号：16340008

Question:

You're given strings J representing the types of stones that are jewels, and S representing the stones you have. Each character in Sis 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:

语言用的是C#

思路在于，创造一个数组以记录某个字母是否在J中出现过，为了方便直接创建了长度为128的布尔数组。

因为直接根据字母的ASCII码作为数组索引，因此查阅布尔值可以达到最快速度。

可以做到J和S两个字符串都只遍历一次。

首先遍历J，以获得字母的出现与否，得到比照数组。

然后遍历S，对S中的每个字母，对照上面的数组对应的布尔值，以判断是否对所求数值操作。

public class Solution {
    public int NumJewelsInStones(string J, string S) {
        int num = 0;
        bool[] letters = new bool[128];
        char[] Jc = J.ToCharArray();
        foreach(char l in Jc) {
            letters[l] = true;
        }
        char[] Sc = S.ToCharArray();
        foreach(char l in Sc) {
            if (letters[l]) 
                num ++;
        }
        return num;
    }
}