[LeetCode] 482. License Key Formatting

You are given a license key represented as a string s that consists of only alphanumeric characters and dashes. The string is separated into n + 1 groups by n dashes. You are also given an integer k.

We want to reformat the string s 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 you should convert all lowercase letters to uppercase.

Return the reformatted license key.

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.

Constraints:

• 1 <= s.length <= 105
• s consists of English letters, digits, and dashes '-'.
• 1 <= k <= 104 

密钥格式化。

有一个密钥字符串 S ，只包含字母，数字以及 '-'（破折号）。其中， N 个 '-' 将字符串分成了 N+1 组。

给你一个数字 K，请你重新格式化字符串，使每个分组恰好包含 K 个字符。特别地，第一个分组包含的字符个数必须小于等于 K，但至少要包含 1 个字符。两个分组之间需要用 '-'（破折号）隔开，并且将所有的小写字母转换为大写字母。

给定非空字符串 S 和数字 K，按照上面描述的规则进行格式化。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/license-key-formatting
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

这是一道字符串的题。我的思路是首先用一个 stringbuilder 把所有不是破折号的char拿出来放在一起。拿出来之后，这里有一个 corner case 需要判断，因为有几个 test case 是只有破折号，没有char的。

其他正常的 case，我们再从右往左遍历，遍历的同时把字母改成大写并放入一个 stack 中，这里同时我们记得要一边数字母的个数一边加破折号。最后我们再把所有内容从stack中都pop出来，输出最后的字符串。

时间O(n)

空间O(n)

Java实现

class Solution {
    public String licenseKeyFormatting(String s, int k) {
        // 收集所有的char
        StringBuilder sb = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (c != '-') {
                sb.append(c);
            }
        }

        int len = sb.length();
        // corner case
        if (len == 0) {
            return "";
        }

        // normal case
        k %= len;
        int kk = k;
        Stack<Character> stack = new Stack<>();
        // 从右往左把char放入stack并改成大写
        for (int i = len - 1; i >= 0; i--) {
            stack.push(Character.toUpperCase(sb.charAt(i)));
            kk--;
            if (kk == 0 && i != 0) {
                stack.push('-');
                kk = k;
            }
        }

        // 输出
        StringBuilder res = new StringBuilder();
        while (!stack.isEmpty()) {
            res.append(stack.pop());
        }
        return res.toString();
    }
}

LeetCode 题目总结