LeetCode771. Jewels and Stones

版权声明：本文为博主原创文章，欢迎转载！转载请保留原博客地址。 https://blog.csdn.net/grllery/article/details/85767244

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

Note:

• S and J will consist of letters and have length at most 50.
• The characters in J are distinct.

---

 题目:计算字符串S中有多少个字符是在字符串J中出现过的.注意:字符串J中的字符均是不同的.

Solution1.对字符串J建立一个map或者set,然后依次判断S中的字符是否在J中.

Solution2:因为S和J中的字符仅有大小写的英文字符,可利用ASCII码表,创建一个大小为128的数组char_num,先统计S中的每个字符出现的次数,然后对J中的每个字符,在数组char_num中查找其对应的值.

#include<unordered_map>
#include<iostream>
using namespace std;

class Solution {
public:
    int numJewelsInStones(string J, string S) {
        unordered_map<char, int> char_num;
        int res = 0;
        for(const auto &c : J){
            char_num[c] = 1;
        }
        
        //判断Stone中的每个元素是否在Jewels中
        for(const auto &c: S){
            auto iter = char_num.find(c);
            if(iter != char_num.end())
                res++;
        }
        return res;
    }
};

class Solution2 {
public:
    int numJewelsInStones(string J, string S) {
        int res = 0;
        //只有大小写字母,对应的ASCII码A:65,a:97.
        int char_num[128] = {0};
        for(const auto &c : S){
            char_num[c]++;
        }
        
        //对于Jewels中的每个元素,在Stone对应的char_num数组中查找
        for(const auto &c: J){
            res += char_num[c];
        }
        return res;
    }
};

int main(int argc, char const *argv[])
{
    Solution sln;
    cout << sln.numJewelsInStones("aA", "aAAbbbb") << endl;
    return 0;
}