问题
A palindrome is a word, phrase, number or other sequence of units that can be read the same way in either direction.
To check whether a word is a palindrome I get the char array of the word and compare the chars. I tested it and it seems to work. However I want to know if it is right or if there is something to improve.
Here is my code:
public class Aufg1 {
public static void main(String[] args) {
String wort = \"reliefpfpfeiller\";
char[] warray = wort.toCharArray();
System.out.println(istPalindrom(warray));
}
public static boolean istPalindrom(char[] wort){
boolean palindrom = false;
if(wort.length%2 == 0){
for(int i = 0; i < wort.length/2-1; i++){
if(wort[i] != wort[wort.length-i-1]){
return false;
}else{
palindrom = true;
}
}
}else{
for(int i = 0; i < (wort.length-1)/2-1; i++){
if(wort[i] != wort[wort.length-i-1]){
return false;
}else{
palindrom = true;
}
}
}
return palindrom;
}
}
回答1:
Why not just:
public static boolean istPalindrom(char[] word){
int i1 = 0;
int i2 = word.length - 1;
while (i2 > i1) {
if (word[i1] != word[i2]) {
return false;
}
++i1;
--i2;
}
return true;
}
Example:
Input is "andna".
i1 will be 0 and i2 will be 4.
First loop iteration we will compare word[0] and word[4]. They're equal, so we increment i1 (it's now 1) and decrement i2 (it's now 3).
So we then compare the n's. They're equal, so we increment i1 (it's now 2) and decrement i2 (it's 2).
Now i1 and i2 are equal (they're both 2), so the condition for the while loop is no longer true so the loop terminates and we return true.
回答2:
You can check if a string is a palindrome by comparing it to the reverse of itself:
public static boolean isPalindrome(String str) {
return str.equals(new StringBuilder(str).reverse().toString());
}
or for versions of Java earlier than 1.5,
public static boolean isPalindrome(String str) {
return str.equals(new StringBuffer().append(str).reverse().toString());
}
EDIT: @FernandoPelliccioni provided a very thorough analysis of the efficiency (or lack thereof) of this solution, both in terms of time and space. If you're interested in the computational complexity of this and other possible solutions to this question, please read it!
回答3:
A concise version, that doesn't involve (inefficiently) initializing a bunch of objects:
boolean isPalindrome(String str) {
int n = str.length();
for( int i = 0; i < n/2; i++ )
if (str.charAt(i) != str.charAt(n-i-1)) return false;
return true;
}
回答4:
Alternatively, recursion.
For anybody who is looking for a shorter recursive solution, to check if a given string satisfies as a palindrome:
private boolean isPalindrome(String s) {
int length = s.length();
if (length < 2) // If the string only has 1 char or is empty
return true;
else {
// Check opposite ends of the string for equality
if (s.charAt(0) != s.charAt(length - 1))
return false;
// Function call for string with the two ends snipped off
else
return isPalindrome(s.substring(1, length - 1));
}
}
OR even shorter, if you'd like:
private boolean isPalindrome(String s) {
int length = s.length();
if (length < 2) return true;
return s.charAt(0) != s.charAt(length - 1) ? false :
isPalindrome(s.substring(1, length - 1));
}
回答5:
Go, Java:
public boolean isPalindrome (String word) {
String myWord = word.replaceAll("\\s+","");
String reverse = new StringBuffer(myWord).reverse().toString();
return reverse.equalsIgnoreCase(myWord);
}
isPalindrome("Never Odd or Even"); // True
isPalindrome("Never Odd or Even1"); // False
回答6:
also a different looking solution:
public static boolean isPalindrome(String s) {
for (int i=0 , j=s.length()-1 ; i<j ; i++ , j-- ) {
if ( s.charAt(i) != s.charAt(j) ) {
return false;
}
}
return true;
}
回答7:
And here a complete Java 8 streaming solution. An IntStream provides all indexes til strings half length and then a comparision from the start and from the end is done.
public static void main(String[] args) {
for (String testStr : Arrays.asList("testset", "none", "andna", "haah", "habh", "haaah")) {
System.out.println("testing " + testStr + " is palindrome=" + isPalindrome(testStr));
}
}
public static boolean isPalindrome(String str) {
return IntStream.range(0, str.length() / 2)
.noneMatch(i -> str.charAt(i) != str.charAt(str.length() - i - 1));
}
Output is:
testing testset is palindrome=true
testing none is palindrome=false
testing andna is palindrome=true
testing haah is palindrome=true
testing habh is palindrome=false
testing haaah is palindrome=true
回答8:
public class Palindromes {
public static void main(String[] args) {
String word = "reliefpfpfeiller";
char[] warray = word.toCharArray();
System.out.println(isPalindrome(warray));
}
public static boolean isPalindrome(char[] word){
if(word.length%2 == 0){
for(int i = 0; i < word.length/2-1; i++){
if(word[i] != word[word.length-i-1]){
return false;
}
}
}else{
for(int i = 0; i < (word.length-1)/2-1; i++){
if(word[i] != word[word.length-i-1]){
return false;
}
}
}
return true;
}
}
回答9:
public class palindrome {
public static void main(String[] args) {
StringBuffer strBuf1 = new StringBuffer("malayalam");
StringBuffer strBuf2 = new StringBuffer("malayalam");
strBuf2.reverse();
System.out.println(strBuf2);
System.out.println((strBuf1.toString()).equals(strBuf2.toString()));
if ((strBuf1.toString()).equals(strBuf2.toString()))
System.out.println("palindrome");
else
System.out.println("not a palindrome");
}
}
回答10:
I worked on a solution for a question that was marked as duplicate of this one. Might as well throw it here...
The question requested a single line to solve this, and I took it more as the literary palindrome - so spaces, punctuation and upper/lower case can throw off the result.
Here's the ugly solution with a small test class:
public class Palindrome {
public static boolean isPalendrome(String arg) {
return arg.replaceAll("[^A-Za-z]", "").equalsIgnoreCase(new StringBuilder(arg).reverse().toString().replaceAll("[^A-Za-z]", ""));
}
public static void main(String[] args) {
System.out.println(isPalendrome("hiya"));
System.out.println(isPalendrome("star buttons not tub rats"));
System.out.println(isPalendrome("stab nail at ill Italian bats!"));
return;
}
}
Sorry that it is kind of nasty - but the other question specified a one-liner.
回答11:
Checking palindrome for first half of the string with the rest, this case assumes removal of any white spaces.
public int isPalindrome(String a) {
//Remove all spaces and non alpha characters
String ab = a.replaceAll("[^A-Za-z0-9]", "").toLowerCase();
//System.out.println(ab);
for (int i=0; i<ab.length()/2; i++) {
if(ab.charAt(i) != ab.charAt((ab.length()-1)-i)) {
return 0;
}
}
return 1;
}
回答12:
I'm new to java and I'm taking up your question as a challenge to improve my knowledge.
import java.util.ArrayList;
import java.util.List;
public class PalindromeRecursiveBoolean {
public static boolean isPalindrome(String str) {
str = str.toUpperCase();
char[] strChars = str.toCharArray();
List<Character> word = new ArrayList<>();
for (char c : strChars) {
word.add(c);
}
while (true) {
if ((word.size() == 1) || (word.size() == 0)) {
return true;
}
if (word.get(0) == word.get(word.size() - 1)) {
word.remove(0);
word.remove(word.size() - 1);
} else {
return false;
}
}
}
}
- If the string is made of no letters or just one letter, it is a palindrome.
- Otherwise, compare the first and last letters of the string.
- If the first and last letters differ, then the string is not a palindrome
- Otherwise, the first and last letters are the same. Strip them from the string, and determine whether the string that remains is a palindrome. Take the answer for this smaller string and use it as the answer for the original string then repeat from 1.
回答13:
Try this out :
import java.util.*;
public class str {
public static void main(String args[])
{
Scanner in=new Scanner(System.in);
System.out.println("ENTER YOUR STRING: ");
String a=in.nextLine();
System.out.println("GIVEN STRING IS: "+a);
StringBuffer str=new StringBuffer(a);
StringBuffer str2=new StringBuffer(str.reverse());
String s2=new String(str2);
System.out.println("THE REVERSED STRING IS: "+str2);
if(a.equals(s2))
System.out.println("ITS A PALINDROME");
else
System.out.println("ITS NOT A PALINDROME");
}
}
回答14:
public boolean isPalindrome(String abc){
if(abc != null && abc.length() > 0){
char[] arr = abc.toCharArray();
for (int i = 0; i < arr.length/2; i++) {
if(arr[i] != arr[arr.length - 1 - i]){
return false;
}
}
return true;
}
return false;
}
回答15:
Another way is using char Array
public class Palindrome {
public static void main(String[] args) {
String str = "madam";
if(isPalindrome(str)) {
System.out.println("Palindrome");
} else {
System.out.println("Not a Palindrome");
}
}
private static boolean isPalindrome(String str) {
// Convert String to char array
char[] charArray = str.toCharArray();
for(int i=0; i < str.length(); i++) {
if(charArray[i] != charArray[(str.length()-1) - i]) {
return false;
}
}
return true;
}
}
回答16:
Here my analysis of the @Greg answer: componentsprogramming.com/palindromes
Sidenote: But, for me it is important to do it in a Generic way. The requirements are that the sequence is bidirectionally iterable and the elements of the sequence are comparables using equality. I don't know how to do it in Java, but, here is a C++ version, I don't know a better way to do it for bidirectional sequences.
template <BidirectionalIterator I>
requires( EqualityComparable< ValueType<I> > )
bool palindrome( I first, I last )
{
I m = middle(first, last);
auto rfirst = boost::make_reverse_iterator(last);
return std::equal(first, m, rfirst);
}
Complexity: linear-time,
If I is RandomAccessIterator: floor(n/2) comparissons and floor(n/2)*2 iterations
If I is BidirectionalIterator: floor(n/2) comparissons and floor(n/2)*2 iterations plus (3/2)*n iterations to find the middle ( middle function )
storage: O(1)
No dymamic allocated memory
回答17:
Recently I wrote a palindrome program which doesn't use StringBuilder. A late answer but this might come in handy to some people.
public boolean isPalindrome(String value) {
boolean isPalindrome = true;
for (int i = 0 , j = value.length() - 1 ; i < j ; i ++ , j --) {
if (value.charAt(i) != value.charAt(j)) {
isPalindrome = false;
}
}
return isPalindrome;
}
回答18:
Using stack, it can be done like this
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String str=in.nextLine();
str.replaceAll("\\s+","");
//System.out.println(str);
Stack<String> stack=new Stack<String>();
stack.push(str);
String str_rev=stack.pop();
if(str.equals(str_rev)){
System.out.println("Palindrome");
}else{
System.out.println("Not Palindrome");
}
}
}
回答19:
public static boolean isPalindrome(String word) {
String str = "";
for (int i=word.length()-1; i>=0; i--){
str = str + word.charAt(i);
}
if(str.equalsIgnoreCase(word)){
return true;
}else{
return false;
}
}
回答20:
Amazing how many different solutions to such a simple problem exist! Here's another one.
private static boolean palindrome(String s){
String revS = "";
String checkS = s.toLowerCase();
String[] checkSArr = checkS.split("");
for(String e : checkSArr){
revS = e + revS;
}
return (checkS.equals(revS)) ? true : false;
}
回答21:
- This implementation works for numbers and strings.
- Since we are not writing anything, so there is no need to convert the string into the character array.
public static boolean isPalindrome(Object obj)
{
String s = String.valueOf(obj);
for(int left=0, right=s.length()-1; left < right; left++,right--)
{
if(s.charAt(left++) != s.charAt(right--))
return false;
}
return true;
}
回答22:
import java.util.Scanner;
public class Palindrom {
public static void main(String []args)
{
Scanner in = new Scanner(System.in);
String str= in.nextLine();
int x= str.length();
if(x%2!=0)
{
for(int i=0;i<x/2;i++)
{
if(str.charAt(i)==str.charAt(x-1-i))
{
continue;
}
else
{
System.out.println("String is not a palindrom");
break;
}
}
}
else
{
for(int i=0;i<=x/2;i++)
{
if(str.charAt(i)==str.charAt(x-1-i))
{
continue;
}
else
{
System.out.println("String is not a palindrom");
break;
}
}
}
}
}
回答23:
private static boolean isPalindrome(String word) {
int z = word.length();
boolean isPalindrome = false;
for (int i = 0; i <= word.length() / 2; i++) {
if (word.charAt(i) == word.charAt(--z)) {
isPalindrome = true;
}
}
return isPalindrome;
}
回答24:
I was looking for a solution that not only worked for palindromes like...
- "Kayak"
- "Madam"
...but as well for...
- "A man, a plan, a canal, Panama!"
- "Was it a car or a cat I saw?"
- "No 'x' in Nixon"
Iterative: This has be proven as a good solution.
private boolean isPalindromeIterative(final String string)
{
final char[] characters =
string.replaceAll("[\\W]", "").toLowerCase().toCharArray();
int iteratorLeft = 0;
int iteratorEnd = characters.length - 1;
while (iteratorEnd > iteratorLeft)
{
if (characters[iteratorLeft++] != characters[iteratorEnd--])
{
return false;
}
}
return true;
}
Recursive. I think this solution shouldn't be much worse than the iterative one. Is a little bit crapy we need to extract the cleaning step out of the method to avoid unnecesary procesing.
private boolean isPalindromeRecursive(final String string)
{
final String cleanString = string.replaceAll("[\\W]", "").toLowerCase();
return isPalindromeRecursiveRecursion(cleanString);
}
private boolean isPalindromeRecursiveRecursion(final String cleanString)
{
final int cleanStringLength = cleanString.length();
return cleanStringLength <= 1 || cleanString.charAt(0) ==
cleanString.charAt(cleanStringLength - 1) &&
isPalindromeRecursiveRecursion
(cleanString.substring(1, cleanStringLength - 1));
}
Reversing: This has been proved as a expensive solution.
private boolean isPalindromeReversing(final String string)
{
final String cleanString = string.replaceAll("[\\W]", "").toLowerCase();
return cleanString.equals(new StringBuilder(cleanString).reverse().toString());
}
All the credits to the guys answering in this post and bringing light to the topic.
回答25:
Considering not letters in the words
public static boolean palindromeWords(String s ){
int left=0;
int right=s.length()-1;
while(left<=right){
while(left<right && !Character.isLetter(s.charAt(left))){
left++;
}
while(right>0 && !Character.isLetter(s.charAt(right))){
right--;
}
if((s.charAt(left++))!=(s.charAt(right--))){
return false;
}
}
return true;
}
———
@Test
public void testPalindromeWords(){
assertTrue(StringExercise.palindromeWords("ece"));
assertTrue(StringExercise.palindromeWords("kavak"));
assertFalse(StringExercise.palindromeWords("kavakdf"));
assertTrue(StringExercise.palindromeWords("akka"));
assertTrue(StringExercise.palindromeWords("??e@@c_--e"));
}
回答26:
Here you can check palindrome a number of String dynamically
import java.util.Scanner;
public class Checkpalindrome {
public static void main(String args[]) {
String original, reverse = "";
Scanner in = new Scanner(System.in);
System.out.println("Enter How Many number of Input you want : ");
int numOfInt = in.nextInt();
original = in.nextLine();
do {
if (numOfInt == 0) {
System.out.println("Your Input Conplete");
}
else {
System.out.println("Enter a string to check palindrome");
original = in.nextLine();
StringBuffer buffer = new StringBuffer(original);
reverse = buffer.reverse().toString();
if (original.equalsIgnoreCase(reverse)) {
System.out.println("The entered string is Palindrome:"+reverse);
}
else {
System.out.println("The entered string is not Palindrome:"+reverse);
}
}
numOfInt--;
} while (numOfInt >= 0);
}
}
回答27:
IMO, the recursive way is the simplest and clearest.
public static boolean isPal(String s)
{
if(s.length() == 0 || s.length() == 1)
return true;
if(s.charAt(0) == s.charAt(s.length()-1))
return isPal(s.substring(1, s.length()-1));
return false;
}
回答28:
here, checking for the largest palindrome in a string, always starting from 1st char.
public static String largestPalindromeInString(String in) {
int right = in.length() - 1;
int left = 0;
char[] word = in.toCharArray();
while (right > left && word[right] != word[left]) {
right--;
}
int lenght = right + 1;
while (right > left && word[right] == word[left]) {
left++;
right--;
}
if (0 >= right - left) {
return new String(Arrays.copyOf(word, lenght ));
} else {
return largestPalindromeInString(
new String(Arrays.copyOf(word, in.length() - 1)));
}
}
回答29:
Code Snippet:
import java.util.Scanner;
class main
{
public static void main(String []args)
{
Scanner sc = new Scanner(System.in);
String str = sc.next();
String reverse = new StringBuffer(str).reverse().toString();
if(str.equals(reverse))
System.out.println("Pallindrome");
else
System.out.println("Not Pallindrome");
}
}
回答30:
import java.util.Collections;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;
public class GetAllPalindromes
{
static Scanner in;
public static void main(String[] args)
{
in = new Scanner(System.in);
System.out.println("Enter a string \n");
String abc = in.nextLine();
Set a = printAllPalindromes(abc);
System.out.println("set is " + a);
}
public static Set<CharSequence> printAllPalindromes(String input)
{
if (input.length() <= 2) {
return Collections.emptySet();
}
Set<CharSequence> out = new HashSet<CharSequence>();
int length = input.length();
for (int i = 1; i < length - 1; i++)
{
for (int j = i - 1, k = i + 1; j >= 0 && k < length; j--, k++)
{
if (input.charAt(j) == input.charAt(k)) {
out.add(input.subSequence(j, k + 1));
} else {
break;
}
}
}
return out;
}
}
**Get All Palindrome in s given string**
Output D:\Java>java GetAllPalindromes Enter a string
Hello user nitin is my best friend wow !
Answer is set is [nitin, nitin , wow , wow, iti]
D:\Java>
来源：https://stackoverflow.com/questions/4138827/check-string-for-palindrome