题目描述

You are given a license key represented as a string S which consists only alphanumeric character and dashes. The string is separated into N+1 groups by N dashes.

Given a number K, we would want to reformat the strings such that each group contains exactly K characters, except for the first group which could be shorter than K, but still must contain at least one character. Furthermore, there must be a dash inserted between two groups and all lowercase letters should be converted to uppercase.

Given a non-empty string S and a number K, format the string according to the rules described above.

Example 1:

Input: S = "5F3Z-2e-9-w", K = 4

Output: "5F3Z-2E9W"

Explanation: The string S has been split into two parts, each part has 4 characters.
Note that the two extra dashes are not needed and can be removed.

Example 2:

Input: S = "2-5g-3-J", K = 2

Output: "2-5G-3J"

Explanation: The string S has been split into three parts, each part has 2 characters except the first part as it could be shorter as mentioned above.

Note:

  1. The length of string S will not exceed 12,000, and K is a positive integer.
  2. String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
  3. String S is non-empty.

解法一

思路

这题换一个说法可以是：将（忽略连接号的）字符串切分为n份长度为K的子串，其中第一份的长度可以少于K；再将这些字串以连接号拼接为一个字符串。

所以这题的关键就是在第一份子串的处理上，这里的第一份就像是除法的余数部分。实际上，我们可以采用从后向前的遍历方法来简单高效地处理这个问题：原字符串从后向前遍历，（忽略原连接号）每遇到K个字符便在其前面添加一个连接号，直到最后一份（实际上是第一份）不足K个字符。这样是不是就像是模拟了一遍整数除法和求余的步骤呢？

当然这里还有一个细节问题要处理：当字符个数整除K的时候，最后添加的那个连接号应该删去（多余）。

Python

class Solution:
    def licenseKeyFormatting(self, S, K):
        """
        :type S: str
        :type K: int
        :rtype: str
        """
        result = ''
        count_k = 0
        for letter in reversed(S):
            if letter != '-':
                result += letter.upper()
                count_k += 1
                if count_k == K:
                    result += '-'
                    count_k = 0
        if len(result) != 0 and result[-1] == '-':
            result = result[:-1]
        return result[::-1]

Java

class Solution {
public:
    string licenseKeyFormatting(string S, int K) {
        string result = "";
        int countK = 0;
        for (int i = S.length() - 1; i >= 0; i--) {
            char letter = S[i];
            if (letter != '-') {
                result += toupper(letter);
                countK++;
                if (countK == K) {
                    result += '-';
                    countK = 0;
                }
            }
        }
        if (result.length() != 0 && result[result.length() - 1] == '-') {
            result.pop_back();
        }
        reverse(result.begin(), result.end());
        return result;
    }
};

C++

public class Solution {
    public String licenseKeyFormatting(String S, int K) {
        StringBuilder result = new StringBuilder();
        int countK = 0;
        for (int i = S.length() - 1; i >= 0; i--) {
            char letter = S.charAt(i);
            if (letter != '-') {
                result.append(letter);
                countK++;
                if (countK == K) {
                    result.append('-');
                    countK = 0;
                }
            }
        }
        if (result.length() != 0 && result.charAt(result.length() - 1) == '-') {
            result.deleteCharAt(result.length() - 1);
        }
        return result.reverse().toString().toUpperCase();
    }
}