LintCode 1038: Jewels And Stones

最新推荐文章于 2021-06-07 11:39:02 发布

原创最新推荐文章于 2021-06-07 11:39:02 发布 · 168 阅读

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本文介绍了一种算法，用于计算给定的石头中，有多少是特定类型的宝石。通过使用两种不同的方法，一种是创建一个映射表来追踪宝石类型，另一种是在宝石字符串中查找石头字符，实现了高效的问题解决。

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
Example 1:

Input: J = “aA”, S = “aAAbbbb”
Output: 3
Example 2:

Input: J = “z”, S = “ZZ”
Output: 0
Notice
S and J will consist of letters and have length at most 50.
The characters in J are distinct.

代码如下：

class Solution {
public:
    /**
     * @param J: the types of stones that are jewels
     * @param S: representing the stones you have
     * @return: how many of the stones you have are also jewels
     */
    int numJewelsInStones(string &J, string &S) {
        vector<int> jewelsTable(255, 0);
        
        int lenJ = J.size();
        int lenS = S.size();
        if (lenJ == 0 || lenS == 0) return 0;
        
        for (int i = 0; i < lenJ; ++i) {
            jewelsTable[J[i]] = 1;
        }
        
        int result = 0;
        for (int i = 0; i < lenS; ++i) {
            result += jewelsTable[S[i]];
        }
        
        return result;
    }
};

网上看到别人的代码，空间效率更好一些：

class Solution {
public:
    /**
     * @param J: the types of stones that are jewels
     * @param S: representing the stones you have
     * @return: how many of the stones you have are also jewels
     */
    int numJewelsInStones(string &J, string &S) {
        // Write your code here
        int count=0;
        for(int i=0;i<S.length();i++)
        {
            if(J.find(S[i])!=string::npos)
            {
                count++;
            }
        }
        return count;
    }
};