LeetCode - 771. Jewels and Stones

题目

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.

常规思路

题目可能有点绕,其实简化一下,就是说有两个字符串JS,两个字符串都是由字母构成的,并且区分大小写,而J里的字母是不重复的,S则是有可能重复的。请找出在S里有多少个字母是与J的字母相同的。

思路很简单,就是遍历S里的每一个字符是否存在于J中,每存在一个就累计一次。代码如下:

class Solution {
    public int numJewelsInStones(String J, String S) {
        int count = 0;
        for (int i = 0; i < S.length(); i++) {
            if (J.indexOf(S.charAt(i)) >= 0)
                count++;
        }
        return count;
    }
}

另辟蹊径

在讨论区里见到了一个很骚的做法,只有一行代码就可以得到想要的结果,如下:

class Solution {
    public int numJewelsInStones(String J, String S) {
        return S.replaceAll("[^" + J + "]", "").length();
    }
}

这种思路是利用了正则表达式,把S里凡是不属于J的字符全部替换成空字符串"",然后剩下的字符串的长度就是最终的答案。

至于这两种思路哪个更好就见仁见智了,下面是两个思路各自花费的时间和内存:

常规思路:

Runtime: 1 ms
Memory Usage: 33.7 MB

特殊思路:

Runtime: 7 ms
Memory Usage: 34.9 MB

相关链接

posted @ 2019-06-08 15:02  雨临Lewis  阅读(...)  评论(... 编辑 收藏