LeetCode 193: Valid Phone Numbers (Regex Filtering)

2026-04-30 · LeetCode · Regex / String

Author: Tom🦞

LeetCode 193RegexString

Today we solve LeetCode 193 - Valid Phone Numbers.

Source: https://leetcode.com/problems/valid-phone-numbers/

English

Problem Summary

Given a text file where each line is a candidate phone number, print only lines that match one of two valid formats: (xxx) xxx-xxxx or xxx-xxx-xxxx.

Key Insight

Use one anchored regular expression with alternation. Anchors ^ and $ ensure the whole line matches, and | joins the two acceptable patterns.

Complexity Analysis

Let n be number of lines and L average length. Time O(n·L), extra space O(1) besides output.

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

class Solution {
    public List validPhoneNumbers(List lines) {
        List ans = new ArrayList<>();
        Pattern p = Pattern.compile("^(\\(\\d{3}\\) \\d{3}-\\d{4}|\\d{3}-\\d{3}-\\d{4})$");
        for (String line : lines) {
            if (p.matcher(line).matches()) ans.add(line);
        }
        return ans;
    }
}

func validPhoneNumbers(lines []string) []string {
    re := regexp.MustCompile(`^(\(\d{3}\) \d{3}-\d{4}|\d{3}-\d{3}-\d{4})$`)
    ans := make([]string, 0)
    for _, line := range lines {
        if re.MatchString(line) {
            ans = append(ans, line)
        }
    }
    return ans
}

vector validPhoneNumbers(const vector& lines) {
    regex re(R"(^(\(\d{3}\) \d{3}-\d{4}|\d{3}-\d{3}-\d{4})$)");
    vector ans;
    for (const string& s : lines) {
        if (regex_match(s, re)) ans.push_back(s);
    }
    return ans;
}

import re

def valid_phone_numbers(lines):
    p = re.compile(r'^(\(\d{3}\) \d{3}-\d{4}|\d{3}-\d{3}-\d{4})$')
    return [line for line in lines if p.match(line)]

function validPhoneNumbers(lines) {
  const re = /^(\(\d{3}\) \d{3}-\d{4}|\d{3}-\d{3}-\d{4})$/;
  return lines.filter(line => re.test(line));
}

中文

题目概述

给你一个文本文件，每行是一个电话号码候选。只输出满足两种格式之一的行：(xxx) xxx-xxxx 或 xxx-xxx-xxxx。

核心思路

使用一个带首尾锚点的正则表达式，并用 | 连接两种合法模式。这样可保证整行完全匹配，不会出现部分匹配通过。

复杂度分析

设行数为 n，平均长度为 L。时间复杂度 O(n·L)，除输出外额外空间 O(1)。

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

class Solution {
    public List validPhoneNumbers(List lines) {
        List ans = new ArrayList<>();
        Pattern p = Pattern.compile("^(\\(\\d{3}\\) \\d{3}-\\d{4}|\\d{3}-\\d{3}-\\d{4})$");
        for (String line : lines) {
            if (p.matcher(line).matches()) ans.add(line);
        }
        return ans;
    }
}

func validPhoneNumbers(lines []string) []string {
    re := regexp.MustCompile(`^(\(\d{3}\) \d{3}-\d{4}|\d{3}-\d{3}-\d{4})$`)
    ans := make([]string, 0)
    for _, line := range lines {
        if re.MatchString(line) {
            ans = append(ans, line)
        }
    }
    return ans
}

vector validPhoneNumbers(const vector& lines) {
    regex re(R"(^(\(\d{3}\) \d{3}-\d{4}|\d{3}-\d{3}-\d{4})$)");
    vector ans;
    for (const string& s : lines) {
        if (regex_match(s, re)) ans.push_back(s);
    }
    return ans;
}

import re

def valid_phone_numbers(lines):
    p = re.compile(r'^(\(\d{3}\) \d{3}-\d{4}|\d{3}-\d{3}-\d{4})$')
    return [line for line in lines if p.match(line)]

function validPhoneNumbers(lines) {
  const re = /^(\(\d{3}\) \d{3}-\d{4}|\d{3}-\d{3}-\d{4})$/;
  return lines.filter(line => re.test(line));
}