482. License Key Formatting - LeetCode

Posted 2020-11-19 okokabcd

Question

482.?License Key Formatting

Solution

思路：字符串转化为char数组，从后遍历，如果是大写字母就转化为小写字母，如果是-就忽略，如果遍历了k个字符(排除-)就追加一个-。

Java实现1：insert版(StringBuilder的append()与insert()效率比较)

public String licenseKeyFormatting(String S, int K) {
    StringBuilder sb = new StringBuilder();
    char[] arr = S.toCharArray();
    int count = 0;
    for (int i = arr.length - 1; i >= 0; i--) {
        char c = arr[i];
        if (c == '-') continue;
        if (count % K == 0) sb.insert(0, '-');
        if (c >= 'a' && c <= 'z') c -= 32;
        sb.insert(0, c);
        count++;
    }
    // return sb.substring(0, sb.length() - 1); // "---" 不通过
    return sb.length() > 0 ? sb.substring(0, sb.length() - 1) : "";
}

Java实现2：append版

public String licenseKeyFormatting(String S, int K) {
    StringBuilder sb = new StringBuilder();
    // char[] arr = S.toCharArray();
    int count = 0;
    for (int i = S.length() - 1; i >= 0; i--) {
        char c = S.charAt(i);
        if (c == '-') continue;
        if (count % K == 0) sb.append('-');//sb.insert(0, '-');
        if (c >= 'a' && c <= 'z') c -= 32;
        sb.append(c);// sb.insert(0, c);
        count++;
    }
    // return sb.substring(0, sb.length() - 1); // "---" 不通过
    return sb.length() > 0 ? sb.reverse().substring(0, sb.length()-1) : "";
}