鏡花水月-什麼是真?什麼是假?

標題: 771. Jewels and Stones

程度:簡單

題目 &範例:

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

這題主要是說使用者會輸入2個變數

一個是 J-->寶石種類，另一個是S-->石頭

大小寫有區分

如範例一: 

J=aA 就是寶石種類有a和A 二種

 

S=aAAbbbb 就是一堆石頭，我們發現這堆石頭中有3個寶石aAA

 

其他bbbb都是石頭，故答案為3個寶石

 

了解題目，程式碼寫起來就挺容易的

 

用for迴圈下去做搜尋S陣列，找到跟J陣列一樣的寶石就count++即可囉!!

 

以下程式碼:

class Solution {
public:
    int numJewelsInStones(string J, string S) {
 int count=0;
        for(int i=0;i<J.size();i++)
        {for(int j=0;j<S.size();j++)
           {
             if(J[i]==S[j]){count++;} 
           }
  }
        return count;
    }
};