Java 不使用内置函数的字符串回文
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String palindrome without using built in functions
提问by Prashant Rai
I want to make program for String palindrome without using built in functions.
我想在不使用内置函数的情况下为字符串回文编写程序。
Below is the code i have tried so far :
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");
}
}
This program is not working properly.I think problem is in in.nextLineand string buffer.
该程序运行不正常。我认为问题出在in.nextLine字符串缓冲区中。
回答by Tamilan C.Periyasamy
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");
}
}
}
Explaination :
说明：
- Given string to convert char array.
Take two point
1st beginning of the string a[0]
2nd one ending of the string a[length-1]
Compare every char in char array
Compare beginning to end and ending begin while looping.
- 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 a[length-1] 的第二个结尾
比较字符数组中的每个字符
循环时比较开始到结束和结束开始。
Palindrom case 1:(length is odd)
回文情况1：（长度为奇数）
"madam" is Palindrom:
“女士”是回文：
a[0] == a[4] m
a[0] == a[4] 米
a[1] == a[3] a
a[1] == a[3] a
a[2] == a[2] d
a[2] == a[2] d
a[3] == a[1] a
a[3] == a[1] a
a[4] == a[0] m
a[4] == a[0] m
Palindrom case 2:(length is even)
回文情况2：（长度是偶数）
"malayalam" is Palindrom:
“马拉雅拉姆语”是回文：
a[0] == a[8] m
a[0] == a[8] 米
a[1] == a[7] a
a[1] == a[7] a
a[2] == a[6] l
a[2] == a[6] l
a[3] == a[5] a
a[3] == a[5] a
a[4] == a[4] y
a[4] == a[4] y
a[5] == a[3] a
a[5] == a[3] a
a[6] == a[2] l
a[6] == a[2] l
a[7] == a[1] a
a[7] == a[1] a
a[8] == a[0] m
a[8] == a[0] 米
Not Palindrom case 3 : Does not match in every chars( beginning to end and ending begin ).
Not Palindrom case 3 : 不匹配每个字符（从开始到结束和结束开始）。
回答by Nilesh Garje
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
回答by Ved Prakash
I saw above all program but it bit hard not understandable to beginners.
我首先看到了所有程序，但初学者很难理解。
Very Simple Program for beginners (to check String Palindrome)
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");
}
}
}
回答by SATYAM SONI
import java.util.*;
class Test{
public static void main(String... as){
int check=0;
Scanner s= new Scanner(System.in);
String str = s.nextLine();
int l=str.length();
for (int i=0;i<l;i++){
if(str.charAt(i)!=str.charAt(l-1-i))
check++;
}
if(check==0)
System.out.println("yes it's palindrome");
else
System.out.println("no it's not.");
}
}
回答by Jitendra Diwaker
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");
}
}
}
回答by gii96
The below given is the simplest way to check if the enteredenter code here string is a palindrome.
下面给出的是检查此处输入的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");
}
}
}
}
回答by SagarS
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");
}
回答by mahitha
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");
}
}
