Leetcode——771. Jewels and Stones

最新推荐文章于 2020-12-27 23:49:46 发布

伍六七AI编程

最新推荐文章于 2020-12-27 23:49:46 发布

阅读量383

点赞数 1

CC 4.0 BY-SA版权

本文为博主原创文章，未经博主允许不得转载。

本文链接：https://blog.youkuaiyun.com/Prepared/article/details/80365328

本文介绍了一种算法问题，即计算给定字符串中属于特定类型字符的数量。提供了两种解决方案，第一种为O(n²)的时间复杂度，通过双重循环实现；第二种为更优的O(n)时间复杂度，利用哈希表进行统计。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

方案一：O（n^2)

class Solution {

    public int numJewelsInStones(String J, String S) {

        int num = 0;

        for(int i = 0; i < J.length(); i++) {

            for(int j = 0; j < S.length(); j++) {

                if(S.charAt(j) == J.charAt(i)) num++;

}

}

        return num;

}

}

Do you have a more effective solution ?

方案二：时间复杂度O（n）

class Solution {

    public int numJewelsInStones(String J, String S) {

        int num = 0;

        Map<String, Integer> map = new HashMap<String, Integer>();

        //将S转成 map<char, int>形式

        for(int i = 0; i < S.length(); i++) {

            if(map.get(String.valueOf(S.charAt(i))) == null) {

                map.put(String.valueOf(S.charAt(i)), 1);

            }else {

                 if(map.get(String.valueOf(S.charAt(i))) != 0 ) {

                     map.put(String.valueOf(S.charAt(i)), map.get(String.valueOf(S.charAt(i))) + 1);

};

}

}

        for(int j = 0; j < J.length(); j++) {

            num = num + map.get(String.valueOf(J.charAt(j)));

}

        return num;

}

}