848. Shifting Letters - Medium
문제
You are given a string s of lowercase English letters and an integer array shifts of the same length.
Call the shift() of a letter, the next letter in the alphabet, (wrapping around so that ‘z’ becomes ‘a’).
- For example, shift(‘a’) = ‘b’, shift(‘t’) = ‘u’, and shift(‘z’) = ‘a’.
Now for each shifts[i] = x, we want to shift the first i + 1 letters of s, x times.
Return the final string after all such shifts to s are applied.
제한사항`
- 1 <= s.length <= 105
- s consists of lowercase English letters.
- shifts.length == s.length
- 0 <= shifts[i] <= 109
입출력 예
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Example 1:
Input: s = "abc", shifts = [3,5,9]
Output: "rpl"
Explanation: We start with "abc".
After shifting the first 1 letters of s by 3, we have "dbc".
After shifting the first 2 letters of s by 5, we have "igc".
After shifting the first 3 letters of s by 9, we have "rpl", the answer.
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Example 2:
Input: s = "aaa", shifts = [1,2,3]
Output: "gfd"
풀이
- String
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class Solution {
public:
string shiftingLetters(string s, vector<int>& shifts) {
// modular 연산을 통해
// 각 인덱스별 최종적으로 shift 해야하는 값 계산
int sum = 0;
for (auto iter = shifts.rbegin() ; iter != shifts.rend() ; ++iter) {
sum = (sum + (*iter % 26)) % 26;
*iter = sum;
}
for (auto i = 0 ; i < s.size() ; ++i) {
auto c = s[i] + shifts[i];
if (c > 'z') {
auto mod = c % 'z';
c = 'a' + mod - 1;
}
s[i] = c;
}
return s;
}
};