问题： 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".

方法： 将J中元素存到set中，遍历S同时统计S中包含的J中元素的数量，最后输出统计的结果。

具体实现：

class JewelsAndStones {    fun numJewelsInStones(J: String, S: String): Int {        val map = mutableSetOf
    
     ()        var count = 0        for (ch in J) {            map.add(ch)        }        for (ch in S) {            if (map.contains(ch)) {                count++            }        }        return count    }}fun main(args: Array
     
      ) {    val J = "aA"    val S = "aAAbbbb"    val jewelsAndStones = JewelsAndStones()    val result = jewelsAndStones.numJewelsInStones(J, S)    println("result: $result")}复制代码
     
    

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