为什么会抛出错误?任何帮助将不胜感激
public class RAWS
{
public String rawsc(String ori)
{
String temp="";
for(int i=0;i<ori.length();i++)
{
char c=ori.charAt(i);
if(((c>=65)&&(c<=90))||((c>=97)&&(c<122)))
temp=c+temp;
}
for(int i=0;i<ori.length();i++)
{
char c=ori.charAt(i);
if(((c>=65)&&(c<=90))||((c>=97)&&(c<122)))
ori.replace(c, temp.charAt(i));
}
for(int i=0;i<ori.length();i++)
{
System.out.println(ori.charAt(i));
}
return(ori);
}
public static void main(String[] args)
{
String str="a,b$c";
RAWS ob=new RAWS();
String new1=ob.rawsc(str);
for(int i=0;i<new1.length();i++)
{
System.out.print(new1.charAt(i)+" ");
}
}
}
编辑:
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 4
at java.lang.String.charAt(String.java:658)
at arraygs.RAWS.rawsc(RAWS.java:22)
at arraygs.RAWS.main(RAWS.java:30)
答案 0 :(得分:1)
有问题的部分是
中的电话temp.charAt(i)
for(int i=0;i<ori.length();i++){
char c=ori.charAt(i);
if(((c>=65)&&(c<=90))||((c>=97)&&(c<122)))
ori.replace(c, temp.charAt(i));
}
字符串temp
的长度可能不是ori
。原因是第一个循环中的if条件
for(int i=0;i<ori.length();i++) {
char c=ori.charAt(i);
if(((c>=65)&&(c<=90))||((c>=97)&&(c<122)))
temp=c+temp;
}
因此,访问i
中的位置temp
(作为第二个循环的一部分)可能会产生java.lang.StringIndexOutOfBoundsException
。
答案 1 :(得分:1)
public class Solution {
public static void main(String[] args) {
System.out.println(reverseString("a,b$c"));
}
/**
* Reverse string with maintaining special character in place
*
* Algorithm:
* 1. create temporary array
* 2. copy all character from original array excluding special character
* 3. reverse the temporary array
* 4. start copying temporary array into original if element is an alphabetic character
* @param input
* @return
*/
public static String reverseString(String input) {
char[] inputArr = input.toCharArray();
char[] tempArr = new char[input.length()];
int i=0;
int j=0;
for (char ch:inputArr){
if(Character.isAlphabetic(ch)){
tempArr[i] = ch;
i++;
}
}
i--;
while(j<i){
char temp = tempArr[i];
tempArr[i]= tempArr[j];
tempArr[j]=temp;
j++;
i--;
}
for(i=0,j=0;i<input.length();i++){
if(Character.isAlphabetic(inputArr[i])){
inputArr[i]= tempArr[j++];
}
}
return new String(inputArr);
}
}
答案 2 :(得分:0)
公共类Ex {
public static void main(String[] args) {
String ss= "Hello@@#+dnksjaf#+43@##@";
char[] c=new char[ss.length()];
String spclCharLessString="";
String spclCharLessStringrev="";
for(int i=0;i<ss.length();i++) {
if(((ss.charAt(i)>='A'&&ss.charAt(i)<='Z')|(ss.charAt(i)>='a'&&ss.charAt(i)<='z')|(ss.charAt(i)>='0'&&ss.charAt(i)<='9'))) {
spclCharLessString+=ss.charAt(i);
}
c[i]=ss.charAt(i);
}
for(int i=spclCharLessString.length()-1;i>=0;i--) {
spclCharLessStringrev+=spclCharLessString.charAt(i);
}
int spclCharSpace=0;
for(int i=0;i<ss.length();i++) {
if(((ss.charAt(i)>='A'&&ss.charAt(i)<='Z')|(ss.charAt(i)>='a'&&ss.charAt(i)<='z')|(ss.charAt(i)>='0'&&ss.charAt(i)<='9'))) {
c[i]=spclCharLessStringrev.charAt(i-spclCharSpace);
}else {
spclCharSpace++;
}
}
System.out.println(spclCharLessStringrev);
for(char c1:c) {
System.out.print(c1);
}
}
}
答案 3 :(得分:0)
使用正则表达式似乎是个好主意。这是我的JavaScript解决方案。
var reverseOnlyLetters = function(S) {
let arr = S.split('')
let regex = /^[a-zA-Z]{1}$/
let i=0,j=arr.length-1;
while(i<j){
if(regex.test(arr[i]) && regex.test(arr[j])){
let temp = arr[i]
arr[i]=arr[j]
arr[j]=temp
i++;j--
}else{
if(!regex.test(arr[i])) i++
if(!regex.test(arr[j])) j--
}
}
return arr.join('')
};
答案 4 :(得分:0)
公共类PracticeJava {
T
}
答案 5 :(得分:0)
str=input("enter any string")
l=[]
s=""
list=list(str)
for i in str:
k=ord(i)
if((k>=48 and k<=57) or (k>=65 and k<=90) or(k>=97 and k<=122)):
l.append(i)
l.reverse()
print(s.join(l))
sreevidhya bontha
答案 6 :(得分:0)
public static void main(String[] args) {
String str = "fed@cb%a!";
char arr[] = new char[str.length()];
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) < 48 || (str.charAt(i) > 57 && str.charAt(i) < 65) || (str.charAt(i) > 90 && str.charAt(i) < 97) || str.charAt(i) > 122)
arr[i] = str.charAt(i);
else
arr[i] = '0';
}
Stack<Character> stack = new Stack<>();
for (int i = 0; i < str.length(); i++) {
stack.push(str.charAt(i));
}
int i=0;
while(!stack.isEmpty()){
char pop = stack.pop();
if (!(pop < 48 || (pop > 57 && pop < 65) || (pop > 90 && pop < 97) || pop > 122)){
arr[i] = pop;
++i;
}
if(arr[i]!='0'){
++i;
}
}
for ( i = 0; i < str.length(); i++) {
System.out.print(arr[i]);
}
}
Time complexity: O(n)
答案 7 :(得分:0)
public static void main(String[] args) {
String a = "ab$cd";
char[] b = a.toCharArray();
int c = b.length;
for (int i = 0; i < c / 2; i++) {
if (Character.isAlphabetic(b[i]) || Character.isDigit(b[i])) {
char temp = b[i];
b[i] = b[c - i - 1];
b[c - i - 1] = temp;
}
}
System.out.println(String.valueOf(b));
}
答案 8 :(得分:0)
公共类MuthuTest { 静态地图列表= new HashMap();
/*
* public static int fact(int n) { if(n>1) return n*fact(n-1); else return 1; }
*/
@SuppressWarnings("deprecation")
public static void main(String[] args) {
String j = "muthu is a good boy.";
int v = 0;
String[] s = j.split(" ");
for (int h = 0; h < s.length; h++) {
String y;
y = s[h].replaceAll("[A-Za-z0-9]", "");
list.put(h, y);
}
// MuthuTest.li(s);
for (int u = s.length - 1; u >= 0; u--) {
if (u == 0) {
s[u] = s[u].replaceAll("[^A-Za-z0-9]", "");
System.out.print(s[u] + list.get(v));
} else {
s[u] = s[u].replaceAll("[^A-Za-z0-9]", "");
System.out.print(s[u] + list.get(v) + " ");
}
v++;
}
}
}
答案 9 :(得分:0)
[Java]仅反转字母而不影响特殊字符的简单方法。
public class StringReverse {
public static void main(String[] args) {
reverseString("T@E$J#A%S");
}
//S@A$JE@T
private static void reverseString(String s){
int len = s.length();
char[] arr = new char[len];
for(int i=0; i<len; i++){
char ch = s.charAt(i);
if(Character.isAlphabetic(ch)){
arr[len-1-i] = ch;
}else{
arr[i] = ch;
}
}
System.out.println(new String(arr));
}
}
答案 10 :(得分:0)
#!/usr/bin/env bash
echo "Choose from 1,2"
read -p "Enter here : " input1
echo "Choose from a,b"
read -p "Enter here : " input2
while true; do
case $input1+$input2 in
1+a )
echo "You have chosen 1 and a"
break;;
1+b )
echo "You have chosen 1 and b"
break;;
2+a )
echo "You have chosen 2 and a"
break;;
2+b )
echo "You have chosen 2 and b"
break;;
*+* )
echo "Invalid input";;
esac
done
答案 11 :(得分:0)
反转字符串而不影响任何特殊字符。请注意,它仅适用于字符串,而不适用于最终将成为字符串数组的字符串组合。
代码:
public class Info {
// Input : str = "Ab,c,de!$" o/p : ed,c,bA!$
public static void main(String[] args) {
String input = "Ab,c,de!$";
char[] inputCharArray = input.toCharArray();
reverseIgnoreSpecialCharacters(inputCharArray);
}
public static void reverseIgnoreSpecialCharacters(char[] charArray) {
int j = charArray.length-1;
int k = 0;
for(int i = charArray.length-1; i>=0; i--) {
if(!(charArray[i] >= 65 && charArray[i] <=90) || !(charArray[i] >= 97 && charArray[i] <=122)) {
charArray[j] = charArray[i];
System.out.print(charArray[j]);
j--;
}
else {
charArray[k] = charArray[i];
System.out.print(charArray[k]);
k++;
}
}
}
}
对于多个字符串,你可以像下面这样:
代码:
public class Info {
// Input : str = "Ab,c,de!$" o/p : ed,c,bA!$
public static void main(String[] args) {
String input = "Ab,c,de!$ Abhi$hek";
String[] inputStringArray = input.split("\\ ");
for(int i = inputStringArray.length-1; i>=0; i--) {
char[] strArray = inputStringArray[i].toCharArray();
reverseIgnoreSpecialCharacters(strArray);
System.out.print(" ");
}
}
public static void reverseIgnoreSpecialCharacters(char[] charArray) {
int j = charArray.length-1;
int k = 0;
for(int i = charArray.length-1; i>=0; i--) {
if(!(charArray[i] >= 65 && charArray[i] <=90) || !(charArray[i] >= 97 && charArray[i] <=122)) {
charArray[j] = charArray[i];
System.out.print(charArray[j]);
j--;
}
else {
charArray[k] = charArray[i];
System.out.print(charArray[k]);
k++;
}
}
}
}