[LeetCode] C++: Simple Question-String 1694. Reformat the phone number

1694. Reformat the phone number

Easy difficulty 6

Give you a phone number in string form  number . number It consists of numbers, spaces  ' ', and dashes  '-' .

Please reformat the phone number as follows.

• First, delete  all spaces and dashes.
• Next, the array from left to right  with each group of three  sub-blocks, until the  remaining four or fewer digits. The remaining numbers will be divided into blocks according to the following rules:
  • 2 numbers: a single block containing 2 numbers.
  • 3 numbers: a single block with 3 numbers.
  • 4 numbers: two blocks each containing 2 numbers.

Finally, connect the blocks with dashes. Note that during the reformatting process,   a block containing only 1 number should not be generated, and a maximum  of two blocks containing 2 numbers should be generated  .

Return the formatted phone number.

 

Example 1:

输入：number = "1-23-45 6"
输出："123-456"
解释：数字是 "123456"
步骤 1：共有超过 4 个数字，所以先取 3 个数字分为一组。第 1 个块是 "123" 。
步骤 2：剩下 3 个数字，将它们放入单个含 3 个数字的块。第 2 个块是 "456" 。
连接这些块后得到 "123-456" 。

Example 2:

输入：number = "123 4-567"
输出："123-45-67"
解释：数字是 "1234567".
步骤 1：共有超过 4 个数字，所以先取 3 个数字分为一组。第 1 个块是 "123" 。
步骤 2：剩下 4 个数字，所以将它们分成两个含 2 个数字的块。这 2 块分别是 "45" 和 "67" 。
连接这些块后得到 "123-45-67" 。

Example 3:

输入：number = "123 4-5678"
输出："123-456-78"
解释：数字是 "12345678" 。
步骤 1：第 1 个块 "123" 。
步骤 2：第 2 个块 "456" 。
步骤 3：剩下 2 个数字，将它们放入单个含 2 个数字的块。第 3 个块是 "78" 。
连接这些块后得到 "123-456-78" 。

Example 4:

输入：number = "12"
输出："12"

Example 5:

输入：number = "--17-5 229 35-39475 "
输出："175-229-353-94-75"

 

prompt:

• 2 <= number.length <= 100
• number Numbers and characters  '-' and  ' ' composition.
• number Contains at least  2  digits in.

 

class Solution {
public:
    string reformatNumber(string number) {
        string buf, res;
        for(auto ch: number){
            if(isdigit(ch)){
                buf.push_back(ch);
            }
        }
        for(int i = 0; i < buf.size(); i++){
            if(i != 0 && i % 3 == 0 && buf.size()-i > 1){
                res.push_back('-');
            }else if(i != 0 && i % 3 == 0 && i == buf.size()-1){
                char ch = res.back();
                res.pop_back();
                res.push_back('-');
                res.push_back(ch);
            }
            res.push_back(buf[i]);
        }
        return res;
    }
};