LeetCode in Kotlin
3602. Hexadecimal and Hexatrigesimal Conversion
Easy
You are given an integer n.
Return the concatenation of the hexadecimal representation of n2 and the hexatrigesimal representation of n3.
A hexadecimal number is defined as a base-16 numeral system that uses the digits 0 – 9 and the uppercase letters A - F to represent values from 0 to 15.
A hexatrigesimal number is defined as a base-36 numeral system that uses the digits 0 – 9 and the uppercase letters A - Z to represent values from 0 to 35.
Example 1:
Input: n = 13
Output: “A91P1”
Explanation:
n2 = 13 * 13 = 169. In hexadecimal, it converts to(10 * 16) + 9 = 169, which corresponds to"A9".n3 = 13 * 13 * 13 = 2197. In hexatrigesimal, it converts to(1 * 362) + (25 * 36) + 1 = 2197, which corresponds to"1P1".- Concatenating both results gives
"A9" + "1P1" = "A91P1".
Example 2:
Input: n = 36
Output: “5101000”
Explanation:
n2 = 36 * 36 = 1296. In hexadecimal, it converts to(5 * 162) + (1 * 16) + 0 = 1296, which corresponds to"510".n3 = 36 * 36 * 36 = 46656. In hexatrigesimal, it converts to(1 * 363) + (0 * 362) + (0 * 36) + 0 = 46656, which corresponds to"1000".- Concatenating both results gives
"510" + "1000" = "5101000".
Constraints:
1 <= n <= 1000
Solution
class Solution {
fun concatHex36(n: Int): String {
var t = n * n
var k: Int
val st = StringBuilder()
var tmp = StringBuilder()
while (t > 0) {
k = t % 16
t = t / 16
if (k <= 9) {
tmp.append(('0'.code + k).toChar())
} else {
tmp.append(('A'.code + (k - 10)).toChar())
}
}
for (i in tmp.length - 1 downTo 0) {
st.append(tmp[i])
}
tmp = StringBuilder()
t = n * n * n
while (t > 0) {
k = t % 36
t = t / 36
if (k <= 9) {
tmp.append(('0'.code + k).toChar())
} else {
tmp.append(('A'.code + (k - 10)).toChar())
}
}
for (i in tmp.length - 1 downTo 0) {
st.append(tmp[i])
}
return st.toString()
}
}