Jewels and Stones(宝石和石头)

1 问题抛出

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.

编码实现

1.第一种解法是用暴力匹配的方法，把S字符串中的每个字符，验证在J字符串中是否出现，每出现一次S中的字符，就把统计的个数加1.代码实现如下：

func numJewelsInStones(J string, S string) int {
  
    count := 0
    
    if J != "" && S != "" {
        for _, s := range S {
            for _, j := range J {
                if s == j {
                    count++
                }
            }
        }    
    }
    
    return count
}

fun main() {
    J := "aA"
	S := "aAAbbbb"
    fmt.Println(numJewelsInStones(J, S))
}

2.从写法上来讲，golang的字符串模块strings提供了一种查找的方法strings.Count(S, v)，代码可以修改为：

func numJewelsInStones(J string, S string) int {
    res := 0
    for _, v := range strings.Split(J, "") {
        res += strings.Count(S, v)
    }
    return res
}

这个算法题的主要思想就是要进行字符匹配，欢迎大家给出其他解法。