55 lines
1.6 KiB
C++
55 lines
1.6 KiB
C++
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// Source : https://leetcode.com/problems/replace-all-digits-with-characters/
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// Author : Hao Chen
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// Date : 2021-05-03
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/*****************************************************************************************************
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*
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* You are given a 0-indexed string s that has lowercase English letters in its even indices and
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* digits in its odd indices.
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*
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* There is a function shift(c, x), where c is a character and x is a digit, that returns the x^th
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* character after c.
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*
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* For example, shift('a', 5) = 'f' and shift('x', 0) = 'x'.
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*
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* For every odd index i, you want to replace the digit s[i] with shift(s[i-1], s[i]).
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*
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* Return s after replacing all digits. It is guaranteed that shift(s[i-1], s[i]) will never exceed
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* 'z'.
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*
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* Example 1:
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*
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* Input: s = "a1c1e1"
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* Output: "abcdef"
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* Explanation: The digits are replaced as follows:
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* - s[1] -> shift('a',1) = 'b'
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* - s[3] -> shift('c',1) = 'd'
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* - s[5] -> shift('e',1) = 'f'
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*
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* Example 2:
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*
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* Input: s = "a1b2c3d4e"
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* Output: "abbdcfdhe"
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* Explanation: The digits are replaced as follows:
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* - s[1] -> shift('a',1) = 'b'
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* - s[3] -> shift('b',2) = 'd'
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* - s[5] -> shift('c',3) = 'f'
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* - s[7] -> shift('d',4) = 'h'
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*
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* Constraints:
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*
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* 1 <= s.length <= 100
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* s consists only of lowercase English letters and digits.
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* shift(s[i-1], s[i]) <= 'z' for all odd indices i.
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******************************************************************************************************/
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class Solution {
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public:
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string replaceDigits(string s) {
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for(int i=0; i<s.size(); i+=2) {
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s[i+1] = s[i] + s[i+1] - '0';
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}
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return s;
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}
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};
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