zoukankan      html  css  js  c++  java
  • [LeetCode] 1945. Sum of Digits of String After Convert

    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.

    字符串转化后的各位数字之和。

    给你一个由小写字母组成的字符串 s ,以及一个整数 k 。

    首先,用字母在字母表中的位置替换该字母,将 s 转化 为一个整数(也就是,'a' 用 1 替换,'b' 用 2 替换,... 'z' 用 26 替换)。接着,将整数 转换 为其 各位数字之和 。共重复 转换 操作 k 次 。

    例如,如果 s = "zbax" 且 k = 2 ,那么执行下述步骤后得到的结果是整数 8 :

    转化:"zbax" ➝ "(26)(2)(1)(24)" ➝ "262124" ➝ 262124
    转换 #1:262124 ➝ 2 + 6 + 2 + 1 + 2 + 4 ➝ 17
    转换 #2:17 ➝ 1 + 7 ➝ 8
    返回执行上述操作后得到的结果整数。

    来源:力扣(LeetCode)
    链接:https://leetcode-cn.com/problems/sum-of-digits-of-string-after-convert
    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。

    这是一道字符串的模拟题,直接根据题意实现即可。因为步骤较多,这道题还是值得一做的。

    时间O(n)

    空间O(n)

    Java实现

     1 class Solution {
     2     public int getLucky(String s, int k) {
     3         StringBuilder sb = new StringBuilder();
     4         for (char c : s.toCharArray()) {
     5             int val = c - 'a' + 1;
     6             sb.append(val);
     7         }
     8 
     9         String numString = sb.toString();
    10         int kk = k;
    11         int sum = 0;
    12         while (kk != 0) {
    13             sum = 0;
    14             for (char c : numString.toCharArray()) {
    15                 int n = c - '0';
    16                 sum += n;
    17             }
    18             kk--;
    19             numString = String.valueOf(sum);
    20         }
    21         return sum;
    22     }
    23 }

    LeetCode 题目总结

  • 相关阅读:
    Node
    HTTP权威指南
    一些乱七八糟的思考积累
    有限状态机
    奇怪问题引用到其它文件夹项目文件方法
    dijkstra+堆优化
    高精度
    BUAA 1301 最短路
    阿里云服务器ECS(Centos8)下安装和配置python3.8
    Cheat Sheet
  • 原文地址:https://www.cnblogs.com/cnoodle/p/15082886.html
Copyright © 2011-2022 走看看