The title
Give you a string s, convert the uppercase letters in the string to the same lowercase letters, and return the new string.
Example 1:
Input: s = "Hello" Output: "Hello"Copy the code
Example 2:
Input: s = "here" Output: "here"Copy the code
Example 3:
Input: s = "LOVELY" Output: "LOVELY"Copy the code
Tip:
1 <= S. length <= 100 s consists of printable characters in the ASCII character setCopy the code
Answer key
Solution one: use Swift language features
Time complexity: O(n)
Spatially complex: O(n)
Code:
func toLowerCase(_ s: String) -> String {
return s.lowercased()
}
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Submission Results:
Solution 2: Write the upper and lower case characters into the dictionary and iterate the query for replacement
Time complexity: O(n)
Spatially complex: O(n)
Code:
func toLowerCase(_ s: String) -> String {
var result: String = s
for item in s {
result = result.replacingOccurrences(of: String(item), with: lower(String(item)))
}
return s.lowercased()
}
private func lower(_ s: String) -> String {
let hash = ["A":"a", "B":"b", "C":"c", "D":"d", "E":"e", "F":"f",
"G":"g", "H":"h", "I":"i", "J":"j", "K":"k", "L":"l",
"M":"m", "N":"n", "O":"o", "P":"p", "Q":"q", "R":"r",
"S":"s", "T":"t", "U":"u", "V":"v", "W":"w", "X":"x",
"Y":"y", "Z":"z"]
if let s1 = hash[s] {
return s1
}
return ""
}
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Submission Results:
Solution 3: use ASCII code
The ASCII code range of uppercase a-z is [65,90]
The ASCII code range for lowercase a-z is [97,122].
Time complexity: O(n)
Spatially complex: O(1)
Code:
func toLowerCase(_ s: String) -> String {
var result: String = ""
for item in s.unicodeScalars {
if item.value >= 65, item.value <= 90 {
result += String(UnicodeScalar(item.value + 32)!)
}else {
result += String(UnicodeScalar(item.value)!)
}
}
return s.lowercased()
}
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Submission Results:
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