TDD LeetCode：771. Jewels and Stones

Rojer Chen's Blog 9月 12, 2018 C# , LeetCode , Programming , TDD Edit

RojerChen.2018.09.12

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

Note:

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

Step1 新增一個類別庫

由於還不知道要如何寫，所以先設定回傳值先固定為 3

public class Algorithm
{
    public int numJewelsInStones(string J, string S)
    {
        return 3;
    }
}

Step2 新增測試案例

依據題目給的測試條件，測試案例如下

[TestMethod]
public void TestMethod1()
{
    Algorithm alg = new Algorithm();
    string firstParam = "aA";
    string secondParam = "aAAbbbb";
    int expected = 3;
    int actual = alg.numJewelsInStones(firstParam, secondParam);

    Assert.AreEqual(expected, actual);
}

執行測試，測試通過

Step3 再新增一個測試案例

依據題目，第二個測試案例如下

[TestMethod]
public void TestMethod2()
{
    Algorithm alg = new Algorithm();
    string firstParam = "z";
    string secondParam = "ZZ";
    int expected = 0;
    int actual = alg.numJewelsInStones(firstParam, secondParam);

    Assert.AreEqual(expected, actual);
}

想當然這次測試不通過了，所以回頭開始調整原本的程式

Step4 調整程式邏輯

調整程式碼邏輯

public int numJewelsInStones(string J, string S)
{
    int counter = 0;
    for (int i = 0; i < J.Length; i++)
    {
        for (int j=0;j<S.Length;j++)
        {
            if (J.Substring(i, 1) == S.Substring(j, 1))
            {
                counter += 1;
            }
        }
    }
    return counter;
}

測試通過。