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
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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 }