【leetcode】1694. Reformat Phone Number

题目如下：

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.

解题思路：很简单的一个题。

代码如下：

class Solution(object):
    def reformatNumber(self, number):
        """
        :type number: str
        :rtype: str
        """
        number = number.replace('-','').replace(' ','')

        remainder = len(number) % 3
        if remainder == 0:
            last_len = 3
            last = number[-3:]
        elif remainder == 1:
            last_len = 4
            last = number[-4:-2] + '-' + number[-2:]
        else:
            last_len = 2
            last = number[-2:]

        if len(number) - last_len <= 0:
            return last
        res = ''
        for i in range(0,len(number) - last_len):
            if i != 0 and i % 3 == 0:
                res += '-'
            res += number[i]
        return res + '-' + last