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ⓁⒸ ‧‧‧ 771. Jewels and Stones

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判斷珠寶與石頭

771. Jewels and Stones 珠寶與石頭

❀ Origin

Problem

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

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Input: J = "aA", S = "aAAbbbb"
Output: 3

Example 2

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Input: J = "z", S = "ZZ"
Output: 0

❀ 翻譯

問題

有一字串 J 代表寶石的種類, 字串 S 代表你擁有的石頭.
字串 S 的每一個字元都代表著你有的一個那種石頭.
你想要知道你的石頭裡有幾個是寶石.

字串 J 的每個字母都保證不同, 而且 J 與 S 的每個字元都保證是字母.
字母有區分大小寫, 所以 “a” 被認定為與 “A” 是不同類型的石頭.

❀ Solution

JavaScript

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/**
 * @param {string} J
 * @param {string} S
 * @return {number}
 */
var numJewelsInStones = function(J, S) {

	/* 依據 J(types of stones that are jewels) 建立一個新的 Set */
	const setJ = new Set([...J]); // Set(2) {"a", "A"}
‍
	/* 依據 S(the stones you have) 建立一個新的 Array */
	const arrayS = S.split(''); // (7) ["a", "A", "A", "b", "b", "b", "b"]

	/* reduce */
	// accumulator: 前一個參數，如果是第一個陣列的話，值是以另外傳入或初始化的值
	// currentValue: 當前變數
	// currentIndex: 當前索引
	// array: 全部陣列

	/* 對 arrayS 跑 reduce,
       使用 Set 的 has(value), 判斷該值是否為Set的成員 */
	const count = arrayS.reduce(
		(accumulator, currentValue, currentIndex, array) => {
			return accumulator + setJ.has(currentValue);
		},
		0 // 傳入reduce accumulator 的初始化值為 0
	);

	return count;
};

Golang

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func numJewelsInStones(J string, S string) int {

	// 建立一個 map 作為 Hash table
	var m = make(map[byte]bool)

	// 將寶石列表 J 的每個值，
	// 作為哈希表的 key 存進去。
	for i := range J {
		m[J[i]] = true
	}

	// 遍歷石頭列表 S ，
	// 逐格去和雜湊表做比對，並記錄總數count
	var count int
	for i := range S {
		if _, ok := m[S[i]]; ok {
			count++
		}
	}

	return count
}