題目
You are given a 0-indexed string s that has lowercase English letters in its even indices and digits in its odd indices.
There is a function shift(c, x), where c is a character and x is a digit, that returns the xth character after c.
For example, shift(‘a’, 5) = ‘f’ and shift(‘x’, 0) = ‘x’. For every odd index i, you want to replace the digit s[i] with shift(s[i-1], s[i]).
Return s after replacing all digits. It is guaranteed that shift(s[i-1], s[i]) will never exceed ‘z’.
Example
Input: s = "a1c1e1"
Output: "abcdef"
Explanation: The digits are replaced as follows:
- s[1] -> shift('a',1) = 'b'
- s[3] -> shift('c',1) = 'd'
- s[5] -> shift('e',1) = 'f'
- 題意: 字串中有數字就要進行位移
解法
func replaceDigits(s string) string {
arr := []byte(s)
for i := 1; i < len(s); i += 2 {
arr[i] = (arr[i-1] - 'a') + arr[i] - '0' + 'a'
}
return string(arr)
}
- 加個測試，看看結果是不是如預期
import (
"github.com/stretchr/testify/assert"
"testing"
)
func Test_replaceDigits(t *testing.T) {
assert.Equal(t, "abcdef", replaceDigits("a1c1e1"))
}