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#yyds干货盘点# 名企真题专题： 回文串
source link: https://blog.51cto.com/u_15488507/5982105
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#yyds干货盘点# 名企真题专题： 回文串
精选 原创1.简述：
描述给定一个字符串，问是否能通过添加一个字母将其变为回文串。
输入描述：一行一个由小写字母构成的字符串，字符串长度小于等于10。
输出描述：输出答案(YES\NO).
示例12.代码实现：
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner input = new Scanner(System.in);
String str ;
boolean re=false;
String op="";
while(input.hasNext()){
str = input.next();
re = false;
for(int i=0;i<str.length();i++){
if(i==0){
op = str.substring(1,str.length());
}else{
if(i == str.length()-1){
op = str.substring(0,str.length()-1);
}else{
op = str.substring(0,i)+str.substring(i+1,str.length());
}
}
re = panduan(op);
if(re){
System.out.println("YES");
break;
}
}
if(re == false){System.out.println("NO");}
}
}
public static boolean panduan(String s){
char[] list = s.toCharArray();
int i=0,j=list.length-1;
while(i<j){
if(list[i] != list[j]){
return false;
}
i++;
j--;
}
return true;
}
}
public class Main{
public static void main(String[] args){
Scanner input = new Scanner(System.in);
String str ;
boolean re=false;
String op="";
while(input.hasNext()){
str = input.next();
re = false;
for(int i=0;i<str.length();i++){
if(i==0){
op = str.substring(1,str.length());
}else{
if(i == str.length()-1){
op = str.substring(0,str.length()-1);
}else{
op = str.substring(0,i)+str.substring(i+1,str.length());
}
}
re = panduan(op);
if(re){
System.out.println("YES");
break;
}
}
if(re == false){System.out.println("NO");}
}
}
public static boolean panduan(String s){
char[] list = s.toCharArray();
int i=0,j=list.length-1;
while(i<j){
if(list[i] != list[j]){
return false;
}
i++;
j--;
}
return true;
}
}
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