125. Valid Palindrome

Easy
​
A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.
Given a string s, return true if it is a palindrome, or false otherwise.
Example 1:
Input: s = "A man, a plan, a canal: Panama"
Output:
true
Explanation:
"amanaplanacanalpanama" is a palindrome.
Example 2:
Input: s = "race a car"
Output:
false
Explanation:
"raceacar" is not a palindrome.
Example 3:
Input: s = " "
Output:
true
Explanation:
s is an empty string "" after removing non-alphanumeric characters.
Since an empty string reads the same forward and backward, it is a palindrome.
Constraints:
  • 1 <= s.length <= 2 * 105
  • s consists only of printable ASCII characters.
​

解題

func isPalindrome(s string) bool {
str := ""
for _, val := range(s) {
if val>=65 && val<=90 {
str+= string(val+32)
} else if ( val>=97 && val<=122 ) || (val>=48 && val<=57) {
str+= string(val)
}
}
start := 0
end := len(str)-1
for start<=end {
if str[start]!=str[end] {
return false
}
start++
end--
}
return true
}
在討論區看到的另一個漂亮解法:
func isPalindrome(s string) bool {
i, j := 0, len(s)-1
for i < j {
if !isValid(s[i]) {
i++
continue
}
if !isValid(s[j]) {
j--
continue
}
if !strings.EqualFold(string(s[i]), string(s[j])) {
return false
}
i++
j--
}
return true
}
​
func isValid(a byte) bool {
if (a >= 'a' && a <= 'z') || (a >= 'A' && a <= 'Z') || (a >= '0' && a <= '9') {
return true
}
return false
}