Question

1694. Reformat Phone Number

中文文档

Description

You are given a phone number as a string number. number consists of digits, spaces ' ', and/or dashes '-'.

You would like to reformat the phone number in a certain manner. Firstly, remove all spaces and dashes. Then, group the digits from left to right into blocks of length 3 until there are 4 or fewer digits. The final digits are then grouped as follows:

• 2 digits: A single block of length 2.
• 3 digits: A single block of length 3.
• 4 digits: Two blocks of length 2 each.

The blocks are then joined by dashes. Notice that the reformatting process should never produce any blocks of length 1 and produce at most two blocks of length 2.

Return _the phone number after formatting._

 

Example 1:

Input: number = "1-23-45 6"
Output: "123-456"
Explanation: The digits are "123456".
Step 1: There are more than 4 digits, so group the next 3 digits. The 1st block is "123".
Step 2: There are 3 digits remaining, so put them in a single block of length 3. The 2nd block is "456".
Joining the blocks gives "123-456".

Example 2:

Input: number = "123 4-567"
Output: "123-45-67"
Explanation: The digits are "1234567".
Step 1: There are more than 4 digits, so group the next 3 digits. The 1st block is "123".
Step 2: There are 4 digits left, so split them into two blocks of length 2. The blocks are "45" and "67".
Joining the blocks gives "123-45-67".

Example 3:

Input: number = "123 4-5678"
Output: "123-456-78"
Explanation: The digits are "12345678".
Step 1: The 1st block is "123".
Step 2: The 2nd block is "456".
Step 3: There are 2 digits left, so put them in a single block of length 2. The 3rd block is "78".
Joining the blocks gives "123-456-78".

 

Constraints:

• 2 <= number.length <= 100
• number consists of digits and the characters '-' and ' '.
• There are at least two digits in number.

Solutions

Solution 1: Simple Simulation

First, according to the problem description, we remove all spaces and hyphens from the string.

Let the current string length be $n$. Then we traverse the string from the beginning, grouping every $3$ characters together and adding them to the result string. We take a total of $n / 3$ groups.

If there is $1$ character left in the end, we form a new group of two characters with the last character of the last group and this character, and add it to the result string. If there are $2$ characters left, we directly form a new group with these two characters and add it to the result string.

Finally, we add hyphens between all groups and return the result string.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.

class Solution:
  def reformatNumber(self, number: str) -> str:
    number = number.replace("-", "").replace(" ", "")
    n = len(number)
    ans = [number[i * 3 : i * 3 + 3] for i in range(n // 3)]
    if n % 3 == 1:
      ans[-1] = ans[-1][:2]
      ans.append(number[-2:])
    elif n % 3 == 2:
      ans.append(number[-2:])
    return "-".join(ans)

class Solution {
  public String reformatNumber(String number) {
    number = number.replace("-", "").replace(" ", "");
    int n = number.length();
    List<String> ans = new ArrayList<>();
    for (int i = 0; i < n / 3; ++i) {
      ans.add(number.substring(i * 3, i * 3 + 3));
    }
    if (n % 3 == 1) {
      ans.set(ans.size() - 1, ans.get(ans.size() - 1).substring(0, 2));
      ans.add(number.substring(n - 2));
    } else if (n % 3 == 2) {
      ans.add(number.substring(n - 2));
    }
    return String.join("-", ans);
  }
}

class Solution {
public:
  string reformatNumber(string number) {
    string s;
    for (char c : number) {
      if (c != ' ' && c != '-') {
        s.push_back(c);
      }
    }
    int n = s.size();
    vector<string> res;
    for (int i = 0; i < n / 3; ++i) {
      res.push_back(s.substr(i * 3, 3));
    }
    if (n % 3 == 1) {
      res.back() = res.back().substr(0, 2);
      res.push_back(s.substr(n - 2));
    } else if (n % 3 == 2) {
      res.push_back(s.substr(n - 2));
    }
    string ans;
    for (auto& v : res) {
      ans += v;
      ans += "-";
    }
    ans.pop_back();
    return ans;
  }
};

func reformatNumber(number string) string {
  number = strings.ReplaceAll(number, " ", "")
  number = strings.ReplaceAll(number, "-", "")
  n := len(number)
  ans := []string{}
  for i := 0; i < n/3; i++ {
    ans = append(ans, number[i*3:i*3+3])
  }
  if n%3 == 1 {
    ans[len(ans)-1] = ans[len(ans)-1][:2]
    ans = append(ans, number[n-2:])
  } else if n%3 == 2 {
    ans = append(ans, number[n-2:])
  }
  return strings.Join(ans, "-")
}

function reformatNumber(number: string): string {
  const cs = [...number].filter(c => c !== ' ' && c !== '-');
  const n = cs.length;
  return cs
    .map((v, i) => {
      if (((i + 1) % 3 === 0 && i < n - 2) || (n % 3 === 1 && n - 3 === i)) {
        return v + '-';
      }
      return v;
    })
    .join('');
}

impl Solution {
  pub fn reformat_number(number: String) -> String {
    let cs: Vec<char> = number
      .chars()
      .filter(|&c| c != ' ' && c != '-')
      .collect();
    let n = cs.len();
    cs.iter()
      .enumerate()
      .map(|(i, c)| {
        if ((i + 1) % 3 == 0 && i < n - 2) || (n % 3 == 1 && i == n - 3) {
          return c.to_string() + &"-";
        }
        c.to_string()
      })
      .collect()
  }
}