[Leetcode] Sum of Digits of String After Convert

김주형·2024년 9월 3일

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참고

You are given a string s consisting of lowercase English letters, and an integer k.

First, convert s into an integer by replacing each letter with its position in the alphabet (i.e., replace 'a' with 1, 'b' with 2, ..., 'z' with 26). Then, transform the integer by replacing it with the sum of its digits. Repeat the transform operation k times in total.

For example, if s = "zbax" and k = 2, then the resulting integer would be 8 by the following operations:

Convert: "zbax" ➝ "(26)(2)(1)(24)" ➝ "262124" ➝ 262124
Transform #1: 262124 ➝ 2 + 6 + 2 + 1 + 2 + 4 ➝ 17
Transform #2: 17 ➝ 1 + 7 ➝ 8
Return the resulting integer after performing the operations described above.

Example 1:

Input: s = "iiii", k = 1
Output: 36
Explanation: The operations are as follows:

Convert: "iiii" ➝ "(9)(9)(9)(9)" ➝ "9999" ➝ 9999
Transform #1: 9999 ➝ 9 + 9 + 9 + 9 ➝ 36
Thus the resulting integer is 36.
Example 2:

Input: s = "leetcode", k = 2
Output: 6
Explanation: The operations are as follows:

Convert: "leetcode" ➝ "(12)(5)(5)(20)(3)(15)(4)(5)" ➝ "12552031545" ➝ 12552031545
Transform #1: 12552031545 ➝ 1 + 2 + 5 + 5 + 2 + 0 + 3 + 1 + 5 + 4 + 5 ➝ 33
Transform #2: 33 ➝ 3 + 3 ➝ 6
Thus the resulting integer is 6.
Example 3:

Input: s = "zbax", k = 2
Output: 8

Constraints:

1 <= s.length <= 100
1 <= k <= 10
s consists of lowercase English letters.

컨텍스트 이해

문자열 변경 후 각 자리수들의 합을 전달해주는 프로그램

기능목록

나의 풀이

// 변경 후 문자열 숫자의 함
// 소문자로 구성된 문자열 s를 

class Solution {
    public int getLucky(String s, int k) {
        int lucky = 0;

        

        return lucky;
    }

    private int convert(String s) {
        int convert = 0;
        for (char index: s.toCharArray()) {
            convert += getNumber(index);
        }
    }

    private void getNumber(char index) {

    }
    
    릿코드를 못 풀었으니 TIL을 기록할게요
    private int transformFirst(int convert) {
        int transformFirst = 0;

        convert

        
    }

    private int transformSecond(int transformFirst) {
        return transformFirst % 10 + (transformFirst / n * 10)
    }
}