2243. 计算字符串的数字和

Java：以 k 为步长进行跳！然后再延伸k

class Solution {
    public String digitSum(String s, int k) {
        while (s.length() > k) {
            int n = s.length();
            StringBuilder sb = new StringBuilder();  // sb存！
            for (int i = 0; i < n; i += k) {
                int sum = 0;
                for (int j = 0; j < k && i + j < n; j++) {
                    sum += s.charAt(i + j) - '0';  // s用于遍历！
                }
                sb.append(sum);
            }
            s = sb.toString();
        }
        return s;
    }
}

Java：自己搓的代码…

class Solution {
    public String digitSum(String s, int k) {
        StringBuilder str = new StringBuilder(s);
        int len = s.length();
        while (len > k) {
            String s1 = new String(str);
            str.delete(0, str.length());
            int a = 0;
            for (int i = 0; i < len; ++i) {
                // a += Integer.valueOf(s1.charAt(i));  // error 入参是String
                a += s1.charAt(i) - '0';
                if((i+1) % k == 0) {
                    str.append(String.valueOf(a));
                    a = 0;
                }
            }
            if (len % k != 0) {
                int b = 0;
                for (int i = len - len % k; i < len; i++) {
                    b += s1.charAt(i) - '0';
                }
                str.append(String.valueOf(b));
            }
            len = str.length();
        }
        return String.valueOf(str);
    }
}