LeetCode 771. Jewels and Stones

最新推荐文章于 2020-12-27 23:49:46 发布

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本文介绍了一个算法问题，即计算给定字符串中属于特定类型字符（宝石）的数量。通过使用C++和Java实现，展示了如何利用set和boolean数组来高效解决这一问题。案例包括输入示例和预期输出。

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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".

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.

//C++
class Solution {
public:
   int numJewelsInStones(string J, string S) {
      //set集合所有元素都会根据元素的键值自动排序，set元素的键值就是实值。set不允许两个元素有相同的键值。
      set<char> setJ(J.begin(), J.end());
      int num = 0;
      for (char s : S) {
         //.count()记录s在setJ中出现一次,在集合中只可能出现1次或0次
         if (setJ.count(s))
            num++;
      }
      return num;
   }
};

//Java
class Solution {
    public int numJewelsInStones(String J, String S) {
        boolean[] dict = new boolean[64];
        for (char c : J.toCharArray()) {
            dict[c - 'A'] = true;
        }
        int ans = 0;
        for (char c : S.toCharArray()) {
            if (dict[c - 'A']) ans++;
        }
        return ans;
    }
}