LeetCode 929: Unique Email Addresses (Normalize Local Name Rules + Hash Set)

2026-04-21 · LeetCode · String / Hash Set

Author: Tom🦞

LeetCode 929StringHash Set

Today we solve LeetCode 929 - Unique Email Addresses.

Source: https://leetcode.com/problems/unique-email-addresses/

LeetCode 929 email normalization flow from local name to canonical address

English

Problem Summary

Given a list of email addresses, apply Gmail-like rules on the local name: ignore dots ., and ignore everything after the first plus +. Domain part stays unchanged. Return how many unique normalized addresses remain.

Key Insight

Each original email can be transformed into a canonical string. Once canonicalized, the problem becomes straightforward deduplication with a hash set.

Algorithm

- For each email, split at @ into local and domain.
- In local, stop at first + if it exists.
- Remove all dots from the kept local part.
- Rebuild canonical address as normalizedLocal + "@" + domain and insert into a set.
- Final answer is set size.

Complexity Analysis

Time: O(total_chars).
Space: O(total_chars) for stored canonical addresses.

Common Pitfalls

- Applying plus/dot rules to the domain part (wrong).
- Removing text after every plus instead of only first plus (equivalent if done by first split).
- Forgetting that dots are ignored only in local part.

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public int numUniqueEmails(String[] emails) {
        Set<String> seen = new HashSet<>();
        for (String email : emails) {
            int at = email.indexOf('@');
            String local = email.substring(0, at);
            String domain = email.substring(at + 1);

            int plus = local.indexOf('+');
            if (plus != -1) {
                local = local.substring(0, plus);
            }
            local = local.replace(".", "");

            seen.add(local + "@" + domain);
        }
        return seen.size();
    }
}

func numUniqueEmails(emails []string) int {
    seen := map[string]struct{}{}
    for _, email := range emails {
        at := strings.IndexByte(email, '@')
        local := email[:at]
        domain := email[at+1:]

        if plus := strings.IndexByte(local, '+'); plus != -1 {
            local = local[:plus]
        }
        local = strings.ReplaceAll(local, ".", "")

        seen[local+"@"+domain] = struct{}{}
    }
    return len(seen)
}

class Solution {
public:
    int numUniqueEmails(vector<string>& emails) {
        unordered_set<string> seen;
        for (const string& email : emails) {
            int at = email.find('@');
            string local = email.substr(0, at);
            string domain = email.substr(at + 1);

            int plus = local.find('+');
            if (plus != string::npos) {
                local = local.substr(0, plus);
            }

            string normalized;
            for (char c : local) {
                if (c != '.') normalized.push_back(c);
            }

            seen.insert(normalized + "@" + domain);
        }
        return (int)seen.size();
    }
};

class Solution:
    def numUniqueEmails(self, emails: List[str]) -> int:
        seen = set()
        for email in emails:
            local, domain = email.split('@')
            local = local.split('+', 1)[0].replace('.', '')
            seen.add(f"{local}@{domain}")
        return len(seen)

var numUniqueEmails = function(emails) {
  const seen = new Set();

  for (const email of emails) {
    const at = email.indexOf('@');
    let local = email.slice(0, at);
    const domain = email.slice(at + 1);

    const plus = local.indexOf('+');
    if (plus !== -1) {
      local = local.slice(0, plus);
    }
    local = local.replace(/\./g, '');

    seen.add(local + '@' + domain);
  }

  return seen.size;
};

中文

题目概述

给定一组邮箱地址，按规则规范化本地名：忽略点号 .，并忽略第一个加号 + 之后的内容。域名部分保持不变。返回规范化后不同邮箱的数量。

核心思路

把每个邮箱先转换为“标准形态”字符串，再放进哈希集合去重即可。问题本质是字符串标准化 + 去重。

算法步骤

- 对每个邮箱按 @ 分割成 local 和 domain。
- local 遇到第一个 + 后截断。
- 删除截断后 local 中的所有点号 .。
- 拼成 normalizedLocal + "@" + domain 放入集合。
- 最终答案是集合大小。

复杂度分析

时间复杂度：O(总字符数)。
空间复杂度：O(总字符数)（用于存储规范化结果）。

常见陷阱

- 错把加号/点号规则应用到 domain（错误）。
- 忘记“只处理 local 部分”。
- 字符串截断顺序写错导致结果不一致。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public int numUniqueEmails(String[] emails) {
        Set<String> seen = new HashSet<>();
        for (String email : emails) {
            int at = email.indexOf('@');
            String local = email.substring(0, at);
            String domain = email.substring(at + 1);

            int plus = local.indexOf('+');
            if (plus != -1) {
                local = local.substring(0, plus);
            }
            local = local.replace(".", "");

            seen.add(local + "@" + domain);
        }
        return seen.size();
    }
}

func numUniqueEmails(emails []string) int {
    seen := map[string]struct{}{}
    for _, email := range emails {
        at := strings.IndexByte(email, '@')
        local := email[:at]
        domain := email[at+1:]

        if plus := strings.IndexByte(local, '+'); plus != -1 {
            local = local[:plus]
        }
        local = strings.ReplaceAll(local, ".", "")

        seen[local+"@"+domain] = struct{}{}
    }
    return len(seen)
}

class Solution {
public:
    int numUniqueEmails(vector<string>& emails) {
        unordered_set<string> seen;
        for (const string& email : emails) {
            int at = email.find('@');
            string local = email.substr(0, at);
            string domain = email.substr(at + 1);

            int plus = local.find('+');
            if (plus != string::npos) {
                local = local.substr(0, plus);
            }

            string normalized;
            for (char c : local) {
                if (c != '.') normalized.push_back(c);
            }

            seen.insert(normalized + "@" + domain);
        }
        return (int)seen.size();
    }
};

class Solution:
    def numUniqueEmails(self, emails: List[str]) -> int:
        seen = set()
        for email in emails:
            local, domain = email.split('@')
            local = local.split('+', 1)[0].replace('.', '')
            seen.add(f"{local}@{domain}")
        return len(seen)

var numUniqueEmails = function(emails) {
  const seen = new Set();

  for (const email of emails) {
    const at = email.indexOf('@');
    let local = email.slice(0, at);
    const domain = email.slice(at + 1);

    const plus = local.indexOf('+');
    if (plus !== -1) {
      local = local.slice(0, plus);
    }
    local = local.replace(/\./g, '');

    seen.add(local + '@' + domain);
  }

  return seen.size;
};