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.
Solution: Simulation
Time complexity: O(klogn)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int getLucky(string s, int k) { int ans = 0; for (char c : s) { int n = c - 'a' + 1; ans += n % 10; ans += n / 10; } while (--k) { int n = ans; ans = 0; while (n) { ans += n % 10; n /= 10; } } return ans; } }; |
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