我想在不使用内置函数的情况下为字符串回文编写程序。
以下是我迄今为止尝试过的代码:
public class Palindrom
{
private static Scanner in;
public static void main(String[] args)
{
String s,str1,str2;
Scanner scan =new Scanner (System.in);
System.out.println("Enter the string");
String s = in.nextLine();
StringBuffer str1 = new StringBuffer();
StringBuffer str2 = new StringBuffer();
str1.reverse();
System.out.println("orignal string="+str2);
System.out.println("reveser string="+str1);
if(String.valueOf(str1).compareTo(String.valueOf(str2))==0)
System.out.println("palindrom");
else
System.out.println("not palindrom");
}
}
此程序工作不正常。我认为问题出在in.nextLine
和字符串缓冲区。
下面给出的是检查此处输入代码字符串是否为回文的最简单方法。
import java.util.Scanner;
public class Strng {
public static void main(String[] args) {
String r = ""; //To store the reverse
Scanner sc = new Scanner(System.in);
System.out.println("Enter the String");
String s = sc.next(); // Entering the string
for(int i= s.length() - 1;i>=0;i--) {
r = r + s.charAt(i);
}
if(r.equals(s)) {
System.out.println("Is a palindrome");
}
else {
System.out.println("Not a palindrome");
}
}
}
public static void palindrome(String str){
Map<Character,Integer> myMap = new HashMap<Character, Integer>();
int characterCounter = 1;
int index=-1;
for(char ch : str.toCharArray()){
index++;
if(str.length()%2!=0 && index==Math.abs(str.length()/2)) continue;
if(myMap.containsKey(ch))
characterCounter = myMap.get(ch)+1;
myMap.put(ch, characterCounter);
}
boolean flag = true;
for(Character myMApVal: myMap.keySet()){
if(myMap.get(myMApVal)%2!=0){
flag=false;
break;
}
}
if(!flag) System.out.println(str+" is not palindrome");
else System.out.println(str+" is palindrome");
}
package practice;
import java.util.Scanner;
public class String_Palindrome {
public static void main(String[] args) {
//String s, str1, str2;
Scanner scanner = new Scanner(System.in);
System.out.println("Enter the string:");
String s = scanner.nextLine();
StringBuffer str1 = new StringBuffer(s);
StringBuffer str2 = new StringBuffer(s);
str1.reverse();
//System.out.println("Original String is:" + str2);
//System.out.println("Reverse String is:" + str1);
if (String.valueOf(str1).compareTo(String.valueOf(str2)) == 0)
System.out.println("Given String is a palindrome");
else
System.out.println("Given String is Not a palindrome");
}
}
下面的代码是检查字符串是否为回文的最简单方法。
public class pal {
static String palindrome(String str) {
char[] array = str.toCharArray();
for (int i = 0; i < array.length / 2; i++) {
if (array[i] != array[array.length - i - 1]) {
return "not a palindrome ";
}
}
return "palindrome ";
}
public static void main(String args[]) {
System.out.println(palindrome("11211"));
}
}
import java.util.Scanner;
public class Palindrom {
static Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
System.out.println("Enter the string");
String inputStr =scanner.next();
System.out.println("Given String = "+inputStr);
char [] charArray=inputStr.toCharArray();
int strlength=(charArray.length)-1;
boolean isPalindrom=true;
for(int count=0; count<charArray.length && strlength >= 0; count++,strlength--){
if(charArray[count]!=charArray[strlength]){
isPalindrom=false;
break;
}
}
if(isPalindrom){
System.out.println("palindrom");
}else{
System.out.println("not palindrom");
}
}
}
解释:
- 给定字符串以转换字符数组
取两点
字符串a[0]的第一个开头
字符串的第二个一端为[长度-1]
比较字符数组中的每个字符
比较循环时的开始到结束和结束开始。
回文情况1:(长度为奇数(
"夫人"是回文:
a[0]==a[4]m
a[1]==a[3]
a[2]==a[2]d
a[3]==a[1]
a[4]==a[0]m
回文情况2:(长度是偶数(
"马拉雅拉姆语"是回文:
a[0]==a[8]m
a[1]==a[7]
a[2]==a[6]l
a[3]==a[5]
a[4]==a[4]y
a[5]==a[3]
a[6]==a[2]l
a[7]==a[1]
a[8]==a[0]米
非回文情况3:不是每个字符都匹配(从开始到结束,从结束到开始(。
import java.util.Scanner;
public class StringPalindrome {
public static void main(String[] args) {
Scanner sc = null;
String str1 = null;
// create object for scanner class
sc = new Scanner(System.in);
if (sc != null) {
System.out.println("Enter First String");
str1 = sc.nextLine();
}
strPalindrom(str1);
}// main
static void strPalindrom(String str) {
// converting string into array
char ch[] = str.toCharArray();
// check string is Palindrom or not
int count = ch.length - 1;
for (int i = 0; i < ch.length; i++, count--) {
if (ch[i] != ch[count]) {
System.out.println("String is not Palindrom");
break;
} else {
if (i == count) {
System.out.println("String is Palindrom");
}
}
} // for
}// strPalindrom method
}// class
我看到了最重要的程序,但它有点难,初学者无法理解。
非常简单的初学者程序(检查字符串回文(
public class Palindrome{
public static void main(String[] args) {
String s = "malayalam"; // enter here your requred String to check Palindrom
String rev = "";
for (int i = s.length() - 1; i >= 0; i--) {
rev = rev + s.charAt(i);
}
if (rev.equals(s)) {
System.out.println("Palindrom");
} else {
System.out.println("Not Palindrom");
}
}
}
public class Palindrom {
public static void main(String[] args) {
String str= "dad";
char[] rev=str.toCharArray();
int i=0;
int j= rev.length-1;
while(i<j){
char temp =rev[i];
rev[i] = rev[j];
rev[j] = temp;
i++;
j--;
}
System.out.println(rev);
System.out.println();
if(str.equals(String.valueOf(rev))){
System.out.println("Yes! It is palindrom");
}else{
System.out.println("No! It is not palindrom");
}
}
}