771. Jewels and Stones
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: 3Example 2:
Input: J = "z", S = "ZZ" Output: 0Note:
S
andJ
will consist of letters and have length at most 50.- The characters in
J
are distinct.
class Solution { public int numJewelsInStones(String J, String S) { int sum = 0; char ss[] = new char[50]; ss = S.toCharArray(); for(char s:ss){ if(J.indexOf(s)!=-1) sum+=1; } return sum; } }这是一道关于字符串的题目,最先想到的做法是把两个字符串拆开两层for循环遍历,在搜索split()方法的时候发现了toCharArray()方法和indexOf(String s) 两个方法 前者可以将字符串转化为字符数组,返回char[] 后者用于判别某字符串是否包含某字串s,若不是返回-1,否则返回其他int
split()方法 stringObj.split(String separator,int limit) stringObj 必选项。要被分解的 String 对象或文字。该对象不会被 split 方法修改。 separator 可选项。字符串或 正则表达式 对象,它标识了分隔字符串时使用的是一个还是多个字符。如果忽 略该选项,返回包含整个字符串的单一元素数组。 limit 可选项。该值用来限制返回数组中的元素个数。one liner解法:
public int numJewelsInStones(String J, String S) { return S.replaceAll("[^" + J + "]", "").length(); }正则表达式
[^>]
表示非>的字符