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[刷题防痴呆] 0482 - 密钥格式化 (License Ke

[刷题防痴呆] 0482 - 密钥格式化 (License Ke

作者: 西出玉门东望长安 | 来源:发表于2021-10-24 01:10 被阅读0次

    题目地址

    https://leetcode.com/problems/license-key-formatting/

    题目描述

    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.
    
    

    思路

    • replace掉所有的"-", 并toUpperCase.
    • 如果当前string的len小于k, 直接返回.
    • 如果len可以被k整除, 则正常按照每k个append.
    • 否则, 需要先取余, 把小于k的个数个char加入stringbuffer中, 之后正常按每k个append即可.

    关键点

    • 注意, 删除掉最后一位多append的"-".

    代码

    • 语言支持:Java
    class Solution {
        public String licenseKeyFormatting(String s, int k) {
            s = s.replace("-", "");
            s = s.toUpperCase();
            StringBuffer sb = new StringBuffer();
            int len = s.length();
            if (len < k) {
                return s;
            }
    
            int index = 0;
            if (len % k == 0) {
                sb.append(s.substring(0, k));
                sb.append("-");
                index = k;
            } else {
                int reminder = len % k;
                sb.append(s.substring(0, reminder));
                sb.append("-");
                index = reminder;
            }
    
            while (index < len) {
                sb.append(s.substring(index, index + k));
                index += k;
                sb.append("-");
            }
            sb.deleteCharAt(sb.length() - 1);
    
            return sb.toString();
        }
    }
    

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