- Valid Palindrome 难度 简单
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.
相关企业 半年内半年 ~ 1年1年 ~ 2年 Facebook|54 微软 Microsoft|11 苹果 Apple|4 彭博 Bloomberg|3
相关标签 Two Pointers String
相似题目 Palindrome Linked List 简单 Valid Palindrome II 简单
- 大思路一样
- 处理很多corner case
- 需要辅助函数
- 2 pointer, check left and right char, move towards middle
- deal with corner cases len <= 1, non letters string
- corner case, "a.", "1p", ".p" (key is which if of 3 if in while is the last exec block)
- Note: alphanumeric include digit
class Solution {
public boolean isPalindrome(String s) {
if (s == null || s.length() == 0){
return false;
}else if (s.length() == 1){
return true;
}
int left = 0;
int right = s.length() - 1;
while(left < right){
if (!isLetterOrDigit(Character.toLowerCase(s.charAt(left)))){
left ++;
continue;
}
if (left == s.length()){ // for non letters string
return true;
}
if (!isLetterOrDigit(Character.toLowerCase(s.charAt(right)))){
right --;
continue;
}
if ( Character.toLowerCase(s.charAt(left)) == Character.toLowerCase(s.charAt(right)) ){
left ++;
right --;
}else{
return false;
}
}
if (left - right == 1){
return true;
}else if (left == right){
return true;
}
return false;
}
public boolean isLetterOrDigit(char c){
return Character.isLetter(c) || Character.isDigit(c);
}
}- faster bc using while inside while loop for some cases
public class Solution {
/**
* @param s: A string
* @return: Whether the string is a valid palindrome
*/
public boolean isPalindrome(String s) {
// write your code here
if (s == null || s.length() == 0){
return true;
}
int left = 0;
int right = s.length() - 1;
while (left < right){
// only check invalid condition
while (left < s.length() && ! isValid(s.charAt(left)) ){
left ++;
}
if (left == s.length()){
return true; //for non letters string
}
while (right >= 0 && !isValid(s.charAt(right))){
right --;
}
if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right)) ){
break;
}else {
left ++;
right --;
}
}
return left >= right;
}
private boolean isValid(char c){
return Character.isLetter(c) || Character.isDigit(c);
}
}- return directly
public class Solution {
/**
* @param s: A string
* @return: Whether the string is a valid palindrome
*/
public boolean isPalindrome(String s) {
// write your code here
if (s == null || s.length() == 0){
return true;
}
int left = 0;
int right = s.length() - 1;
while (left < right){
// only check invalid condition
while (left < right && ! isValid(s.charAt(left)) ){
left ++;
}
while (left < right && !isValid(s.charAt(right))){
right --;
}
if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right)) ){
return false;
}else {
left ++;
right --;
}
}
return true;
}
private boolean isValid(char c){
return Character.isLetter(c) || Character.isDigit(c);
}
}class Solution:
def isPalindrome(self, s: str) -> bool:
if not s:
return False
left, right = 0, len(s) - 1
while left < right:
while left < right and not s[left].isalpha() and not s[left].isdigit():
left += 1
while left < right and not s[right].isalpha() and not s[right].isdigit():
right -= 1
if s[left].lower() == s[right].lower():
left += 1
right -= 1
else:
return False
return True