[leetcode] 771. Jewels and Stones

最新推荐文章于 2023-09-14 23:23:26 发布

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最新推荐文章于 2023-09-14 23:23:26 发布

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本文介绍了一种使用C++ STL中的set容器解决宝石计数问题的方法。通过将宝石类型字符串J中的字符存入set，再遍历石头字符串S检查每个字符是否存在于set中，从而高效计算出S中有多少字符属于J。文中提供了具体的实现代码。

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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.

思路：

1.其实可以运用一次嵌套循环，来判断S中的字符串是否在J中，但是这样的话，复杂度太大。

2.结合C++的STL，set容器，先将J中的字符放进set中，再对S进行遍历，判断S中的字符是否在这个set中，用到find函数。

3.这个find函数比我们自己进行遍历的效率大很多。

4.我们应该熟练STL的用法。

代码：C++

class Solution {
public:
    int numJewelsInStones(string J, string S) {
        set<char> c;
        for(int i = 0; i < J.size(); i++){
            c.insert(J[i]);
        }
        int count = 0;
        for(int i = 0; i < S.size(); i++){
            if(c.find(S[i]) != c.end()){
                count++;
            }
        }
        return count;
    }
};