929. Unique Email Address

最新推荐文章于 2019-09-19 04:18:33 发布

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最新推荐文章于 2019-09-19 04:18:33 发布

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分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/PIG_CODE/article/details/83830186

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If you add a plus ('+') in the local name, everything after the first plus sign will be ignored. This allows certain emails to be filtered, for example m.y+name@email.com will be forwarded to my@email.com. (Again, this rule does not apply for domain names.)

It is possible to use both of these rules at the same time.

Given a list of emails, we send one email to each address in the list. How many different addresses actually receive mails?

Example 1:

Input: ["test.email+alex@leetcode.com","test.e.mail+bob.cathy@leetcode.com","testemail+david@lee.tcode.com"]
Output: 2
Explanation: "testemail@leetcode.com" and "testemail@lee.tcode.com" actually receive mails

本菜鸟代码：

class Solution {
    public int numUniqueEmails(String[] emails) {
        HashSet stringSet = new HashSet();
        String s = null;
        for(int i = 0;i<emails.length;i++){
            s = "";
            int flagplus = 0;
            int flagat = 0;
            for(int j = 0;j<emails[i].length();j++){
                char ss = emails[i].charAt(j);
                if(ss!='@'&&flagplus==0){
                    if(ss!='.'&&ss!='+')  s+=ss;
                    else{
                        if(ss=='+') flagplus = 1;
                    }
                }else{
                    if(ss=='@'||flagat==1) {
                        flagat = 1;
                        s+=ss;
                    }
                }
            }
            stringSet.add(s);
        }
        return stringSet.size();
    }
}

我的思路是：
对于‘+’ 和‘@’设置两个flag，如果遇到‘+’，就一直不添加字符，直到遇到了‘@’。在‘@’以后都直接添加字符。为了统计不重复的字符串，我用了HashSet来存储。最后返回HashSet的大小。

官方的答案是：

class Solution {
    public int numUniqueEmails(String[] emails) {
        Set<String> seen = new HashSet();
        for (String email: emails) {
            int i = email.indexOf('@');
            String local = email.substring(0, i);
            String rest = email.substring(i);
            if (local.contains("+")) {
                local = local.substring(0, local.indexOf('+'));
            }
            local = local.replaceAll(".", "");
            seen.add(local + rest);
        }

        return seen.size();
    }
}

官方的答案非常清楚。但是用到了很多子函数：indexOf , contains, substring,replaceAll.