您好我一直在做这个java程序,我应该输入一个字符串并输出可以找到的最长的回文。 但我的程序只输出最长回文的第一个字母..我非常需要你的帮助..谢谢!
应该是:
INPUT:abcdcbbcdeedcba 输出:bcdeedcb 有两个回文串:bcdcb和bcdeedcb
但是当我输入时:abcdcbbcdeedcba 输出:b
import javax.swing.JOptionPane;
public class Palindrome5
{ public static void main(String args[])
{ String word = JOptionPane.showInputDialog(null, "Input String : ", "INPUT", JOptionPane.QUESTION_MESSAGE);
String subword = "";
String revword = "";
String Out = "";
int size = word.length();
boolean c;
for(int x=0; x<size; x++)
{ for(int y=x+1; y<size-x; y++)
{ subword = word.substring(x,y);
c = comparisonOfreverseword(subword);
if(c==true)
{
Out = GetLongest(subword);
}
}
}
JOptionPane.showMessageDialog(null, "Longest Palindrome : " + Out, "OUTPUT", JOptionPane.PLAIN_MESSAGE);
}
public static boolean comparisonOfreverseword(String a)
{ String rev = "";
int tempo = a.length();
boolean z=false;
for(int i = tempo-1; i>=0; i--)
{
char let = a.charAt(i);
rev = rev + let;
}
if(a.equalsIgnoreCase(rev))
{
z=true;
}
return(z);
}
public static String GetLongest(String sWord)
{
int sLength = sWord.length();
String Lpalindrome = "";
int storage = 0;
if(storage<sLength)
{
storage = sLength;
Lpalindrome = sWord;
}
return(Lpalindrome);
}
}
答案 0 :(得分:1)
修改程序..该程序将提供正确的输出
package pract1;
import javax.swing.JOptionPane;
public class Palindrome5
{
public static void main(String args[])
{
String word = JOptionPane.showInputDialog(null, "Input String : ", "INPUT", JOptionPane.QUESTION_MESSAGE);
String subword = "";
String revword = "";
String Out = "";
int size = word.length();
boolean c;
String Lpalindrome = "";
int storage=0;
String out="";
for(int x=0; x<size; x++)
{ for(int y=x+1; y<=size; y++)
{ subword = word.substring(x,y);
c = comparisonOfreverseword(subword);
if(c==true)
{
int sLength = subword.length();
if(storage<sLength)
{
storage = sLength;
Lpalindrome = subword;
out=Lpalindrome;
}
}
}
}
JOptionPane.showMessageDialog(null, "Longest Palindrome : " + out, "OUTPUT", JOptionPane.PLAIN_MESSAGE);
}
public static boolean comparisonOfreverseword(String a)
{ String rev = "";
int tempo = a.length();
boolean z=false;
for(int i = tempo-1; i>=0; i--)
{
char let = a.charAt(i);
rev = rev + let;
}
if(a.equalsIgnoreCase(rev))
{
z=true;
}
return(z);
}
}
答案 1 :(得分:0)
你有两个错误:
1
for(int y=x+1; y<size-x; y++)
应该是
for(int y=x+1; y<size; y++)
因为你仍然希望一直走到字符串的末尾。使用前一个循环,因为x在整个循环中增加,所以子串大小会在整个循环中减少(通过从末尾删除x个字符)。
2
您没有存储到目前为止找到的最长的字符串或其长度。代码
int storage = 0;
if(storage<sLength) {
storage = sLength;
...
说'如果新字符串长于零个字符,那么我将假设它是迄今为止找到的最长字符串并将其作为LPalindrome返回'。这没有任何帮助,因为我们之前可能已经找到了更长的回文。
如果是我,我会制作一个静态变量(例如longestSoFar)来保持到目前为止找到的最长的回文(最初是空的)。对于每个新的回文,检查新的回文是否长于最长的时间。如果更长,则将其分配给longestSoFar。然后在最后,显示longestSoFar。
一般情况下,如果你在程序中“记住”某些内容时遇到问题(例如以前看过的值),你必须考虑静态存储一些内容,因为一旦方法完成就会忘记局部变量。
答案 2 :(得分:0)
公共类LongestPalindrome {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
String S= "abcdcba";
printLongestPalindrome(S);
}
public static void printLongestPalindrome(String S)
{
int maxBack=-1;
int maxFront = -1;
int maxLength=0;
for (int potentialCenter = 0 ; potentialCenter < S.length();potentialCenter ++ )
{
int back = potentialCenter-1;
int front = potentialCenter + 1;
int longestPalindrome = 0;
while(back >=0 && front<S.length() && S.charAt(back)==S.charAt(front))
{
back--;
front++;
longestPalindrome++;
}
if (longestPalindrome > maxLength)
{
maxLength = longestPalindrome+1;
maxBack = back + 1;
maxFront = front;
}
back = potentialCenter;
front = potentialCenter + 1;
longestPalindrome=0;
while(back >=0 && front<S.length() && S.charAt(back)==S.charAt(front))
{
back--;
front++;
longestPalindrome++;
}
if (longestPalindrome > maxLength)
{
maxLength = longestPalindrome;
maxBack = back + 1;
maxFront = front;
}
}
if (maxLength == 0) System.out.println("There is no Palindrome in the given String");
else{
System.out.println("The Longest Palindrome is " + S.substring(maxBack,maxFront) + "of " + maxLength);
}
}
}
答案 3 :(得分:0)
我有自己的方法以随机单词获得最长回文。检查一下
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println(longestPalSubstr(in.nextLine().toLowerCase()));
}
static String longestPalSubstr(String str) {
char [] input = str.toCharArray();
Set<CharSequence> out = new HashSet<CharSequence>();
int n1 = str.length()-1;
for(int a=0;a<=n1;a++)
{
for(int m=n1;m>a;m--)
{
if(input[a]==input[m])
{
String nw = "",nw2="";
for (int y=a;y<=m;y++)
{
nw=nw+input[y];
}
for (int t=m;t>=a;t--)
{
nw2=nw2+input[t];
}
if(nw2.equals(nw))
{
out.add(nw);
break;
}
}
}
}
int a = out.size();
int maxpos=0;
int max=0;
Object [] s = out.toArray();
for(int q=0;q<a;q++)
{
if(max<s[q].toString().length())
{
max=s[q].toString().length();
maxpos=q;
}
}
String output = "longest palindrome is : "+s[maxpos].toString()+" and the lengths is : "+ max;
return output;
}
此方法将返回最大长度回文及其长度。这是我尝试并获得答案的一种方式。并且此方法无论其是奇数长度还是偶数长度都将运行。
答案 4 :(得分:0)
这是我获得最长回文的方法。这将返回长度和回文词
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println(longestPalSubstr(in.nextLine().toLowerCase()));
}
static String longestPalSubstr(String str) {
char [] input = str.toCharArray();
Set<CharSequence> out = new HashSet<CharSequence>();
int n1 = str.length()-1;
for(int a=0;a<=n1;a++)
{
for(int m=n1;m>a;m--)
{
if(input[a]==input[m])
{
String nw = "",nw2="";
for (int y=a;y<=m;y++)
{
nw=nw+input[y];
}
for (int t=m;t>=a;t--)
{
nw2=nw2+input[t];
}
if(nw2.equals(nw))
{
out.add(nw);
break;
}
}
}
}
int a = out.size();
int maxpos=0;
int max=0;
Object [] s = out.toArray();
for(int q=0;q<a;q++)
{
if(max<s[q].toString().length())
{
max=s[q].toString().length();
maxpos=q;
}
}
String output = "longest palindrome is : "+s[maxpos].toString()+" and the lengths is : "+ max;
return output;
}
此方法将返回最大长度回文及其长度。这是我尝试并获得答案的一种方式。并且此方法无论其是奇数长度还是偶数长度都将运行。
答案 5 :(得分:0)
public class LongestPalindrome {
public static void main(String[] args) {
HashMap<String, Integer> result = findLongestPalindrome("ayrgabcdeedcbaghihg123444456776");
result.forEach((k, v) -> System.out.println("String:" + k + " Value:" + v));
}
private static HashMap<String, Integer> findLongestPalindrome(String str) {
int i = 0;
HashMap<String, Integer> map = new HashMap<String, Integer>();
while (i < str.length()) {
String alpha = String.valueOf(str.charAt(i));
if (str.indexOf(str.charAt(i)) != str.lastIndexOf(str.charAt(i))) {
String pali = str.substring(i, str.lastIndexOf(str.charAt(i)) + 1);
if (isPalindrome(pali)) {
map.put(pali, pali.length());
i = str.lastIndexOf(str.charAt(i));
}
}
i++;
}
return map;
}
public static boolean isPalindrome(String input) {
for (int i = 0; i <= input.length() / 2; i++) {
if (input.charAt(i) != input.charAt(input.length() - 1 - i)) {
return false;
}
}
return true;
}
}
这种方法很简单。
输出:
字符串:abcdeedcba值:10
字符串:4444值:4
字符串:6776值:4
字符串:ghihg值:5