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.
题解:
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