LeetCode125. Valid Palindrome
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2024-03-06 21:38:14
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125.Valid Palindrome
Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
For example,
“A man, a plan, a canal: Panama” is a palindrome.
“race a car” is not a palindrome.
Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.
For the purpose of this problem, we define empty string as valid palindrome.
方法一:超时
public boolean isPalindrome(String s) {
if(s.length() == 0) return true;
List<Character> a = new ArrayList<>();
List<Character> b = new ArrayList<>();
for(int i = 0; i < s.length(); i++){
if(s.charAt(i) >= 'a' && s.charAt(i) <= 'z' ||s.charAt(i) >= 'A' && s.charAt(i) <= 'Z'|| s.charAt(i) >= '0' && s.charAt(i) <= '9'){
a.add(s.charAt(i));
}
}
for(int i = 0; i < a.size(); i++){
b.add(0, a.get(i));
}
if(String.valueOf(a).equalsIgnoreCase(String.valueOf(b))){
return true;
}
return false;
}
方法二:
设置一首一尾两个指针,每次比较两个指针指向的字符是否相等,若相等,则首指针向后移一个位置,尾指针向前移一个位置。
public boolean isPalindrome2(String s) {
if (s.isEmpty()) {
return true;
}
int head = 0, tail = s.length() - 1;
char cHead, cTail;
while(head <= tail) {
cHead = s.charAt(head);
cTail = s.charAt(tail);
if (!Character.isLetterOrDigit(cHead)) {
head++;
} else if(!Character.isLetterOrDigit(cTail)) {
tail--;
} else {
if (Character.toLowerCase(cHead) != Character.toLowerCase(cTail)) {
return false;
}
head++;
tail--;
}
}
return true;
}
方法三:
这个方法用到了正则表达式,并用到了StringBuffer类自带的reverse函数。
public boolean isPalindrome3(String s) {
String actual = s.replaceAll("[^A-Za-z0-9]", "").toLowerCase();
String rev = new StringBuffer(actual).reverse().toString();
return actual.equals(rev);
}
注意:
字符串比较应该用equals()方法,不能用 == 号。