输入一个正整数的字符串，输出与它最接近的对称数字(不包括它自己)的字符串
注1: 输入字符串的长度最多不会超过18
注2: 当大于输入数字和小于输入数字的对称数字与输入数字距离相同时，取小的数字作为答案
[编程题]找出最接近的对称数字
最接近的数一定在这三个中产生，首先自身截半，取左边部分，记得half，数据1为half对称复制（如果对称复制后和原数相同则排除掉，2，half-1对称复制，half+1对称复制，另外要注意处理掉加1减1导致位数变化的情况
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;
public class Main {
public static void main(String[] args) {
Main da = new Main();
Scanner scanner = new Scanner(System.in);
String s = scanner.next();
String duichen = da.duichen(s);
System.out.println(duichen);
}
/**
* 判断一个字符串是否对称
* @param s 字符串
* @return 是否对称
*/
public boolean isRever(String s) {
char[] arr = s.toCharArray();
//以0结尾肯定不对称
if (arr[arr.length - 1] == '0') {
return false;
}
String s2 = new StringBuilder(s).reverse().toString();
return s.equals(s2);
}
public String duichen(String s) {
Set<Long> tmpSet = new HashSet<Long>();
long data = Long.parseLong(s);
char[] arr = s.toCharArray();
if (arr.length == 1) {
long res = data == 0 ? 0 : data - 1;
return String.valueOf(res);
} else {
//奇数位
boolean isJiSu = arr.length%2 != 0 ? true: false;// 是否是奇数位
boolean isFrontChange = isJiSu, isAfterChange=isJiSu;
int middle = isJiSu ? 1 : 0;// 如果是奇数位，加上中间数，偶数位不加
String halfStr = s.substring(0, arr.length / 2 + middle);
long frontHalf = Long.parseLong(halfStr) - 1;
// 如果是10，1000，100111之类，减一后位数变化
boolean isChange = is100(halfStr);
String frontHalfStr = String.valueOf(frontHalf);
if (isChange) {
isFrontChange = !isJiSu;//位数发生变化，奇变偶偶变奇
//偶变奇要补中间9;
if (isFrontChange) {
frontHalfStr = frontHalfStr+"9";
}
}
String frontRevStr = copyRever(frontHalfStr, isFrontChange);
tmpSet.add(Long.parseLong(frontRevStr));
long afterHalf = Long.parseLong(halfStr) + 1;
//如果是99，9999， 99999之类， 加1后位数变化 99 9
isChange = is99(halfStr);//10 : 101
String afterHalfStr = String.valueOf(afterHalf);
if (isChange) {
isAfterChange = !isJiSu;
if (isAfterChange) {
afterHalfStr = afterHalfStr.substring(0, afterHalfStr.length()-1);//减一位0
}
}
String afterRevStr = copyRever(afterHalfStr, isAfterChange);
tmpSet.add(Long.parseLong(afterRevStr));
if (!isRever(s)) {
String revStr = copyRever(halfStr, isJiSu);
tmpSet.add(Long.parseLong(revStr));
}
}
String tmpKey = "";
long tmp = Long.parseLong(s);
for (Long cur : tmpSet) {
long abs = Math.abs(Long.parseLong(s) - cur);
if (tmp > abs) {
tmp = abs;
tmpKey = cur.toString();
}
}
return tmpKey;
}
private boolean is99(String half) {
long halfInt = Long.parseLong(half) + 1;
return String.valueOf(halfInt).length() != half.length();
}
private boolean is100(String half) {
if (half.equals("1")) {
return true;
}
long halfInt = Long.parseLong(half) - 1;
return String.valueOf(halfInt).length() != half.length();
}
public String copyRever(String half, boolean isMiddle) {
if (isMiddle) {
String tmp = half.substring(0, half.length() - 1);
String rev = new StringBuilder(tmp).reverse().toString();
return half + rev;
} else {
String rev = new StringBuilder(half).reverse().toString();
return half + rev;
}
}
}
import java.util.Scanner;
import java.util.Set;
public class Main {
public static void main(String[] args) {
Main da = new Main();
Scanner scanner = new Scanner(System.in);
String s = scanner.next();
String duichen = da.duichen(s);
System.out.println(duichen);
}
/**
* 判断一个字符串是否对称
* @param s 字符串
* @return 是否对称
*/
public boolean isRever(String s) {
char[] arr = s.toCharArray();
//以0结尾肯定不对称
if (arr[arr.length - 1] == '0') {
return false;
}
String s2 = new StringBuilder(s).reverse().toString();
return s.equals(s2);
}
public String duichen(String s) {
Set<Long> tmpSet = new HashSet<Long>();
long data = Long.parseLong(s);
char[] arr = s.toCharArray();
if (arr.length == 1) {
long res = data == 0 ? 0 : data - 1;
return String.valueOf(res);
} else {
//奇数位
boolean isJiSu = arr.length%2 != 0 ? true: false;// 是否是奇数位
boolean isFrontChange = isJiSu, isAfterChange=isJiSu;
int middle = isJiSu ? 1 : 0;// 如果是奇数位，加上中间数，偶数位不加
String halfStr = s.substring(0, arr.length / 2 + middle);
long frontHalf = Long.parseLong(halfStr) - 1;
// 如果是10，1000，100111之类，减一后位数变化
boolean isChange = is100(halfStr);
String frontHalfStr = String.valueOf(frontHalf);
if (isChange) {
isFrontChange = !isJiSu;//位数发生变化，奇变偶偶变奇
//偶变奇要补中间9;
if (isFrontChange) {
frontHalfStr = frontHalfStr+"9";
}
}
String frontRevStr = copyRever(frontHalfStr, isFrontChange);
tmpSet.add(Long.parseLong(frontRevStr));
long afterHalf = Long.parseLong(halfStr) + 1;
//如果是99，9999， 99999之类， 加1后位数变化 99 9
isChange = is99(halfStr);//10 : 101
String afterHalfStr = String.valueOf(afterHalf);
if (isChange) {
isAfterChange = !isJiSu;
if (isAfterChange) {
afterHalfStr = afterHalfStr.substring(0, afterHalfStr.length()-1);//减一位0
}
}
String afterRevStr = copyRever(afterHalfStr, isAfterChange);
tmpSet.add(Long.parseLong(afterRevStr));
if (!isRever(s)) {
String revStr = copyRever(halfStr, isJiSu);
tmpSet.add(Long.parseLong(revStr));
}
}
String tmpKey = "";
long tmp = Long.parseLong(s);
for (Long cur : tmpSet) {
long abs = Math.abs(Long.parseLong(s) - cur);
if (tmp > abs) {
tmp = abs;
tmpKey = cur.toString();
}
}
return tmpKey;
}
private boolean is99(String half) {
long halfInt = Long.parseLong(half) + 1;
return String.valueOf(halfInt).length() != half.length();
}
private boolean is100(String half) {
if (half.equals("1")) {
return true;
}
long halfInt = Long.parseLong(half) - 1;
return String.valueOf(halfInt).length() != half.length();
}
public String copyRever(String half, boolean isMiddle) {
if (isMiddle) {
String tmp = half.substring(0, half.length() - 1);
String rev = new StringBuilder(tmp).reverse().toString();
return half + rev;
} else {
String rev = new StringBuilder(half).reverse().toString();
return half + rev;
}
}
}
发表于 2022-07-13 01:14:27
回复(0)
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
String num = sc.next();
int chang = num.length();
char[] ch =num.toCharArray();
if(chang==1) System.out.println(Integer.parseInt(num)-1);
else if(!sw(num)){
int left = 0;
int right = num.length()-1;
while(left<=right){
ch[right] = ch[left];
left++;
right--;
}
System.out.println(String.valueOf(ch));
}
else{
int i=-1;
int j=1;
int nu = Integer.parseInt(num);
while(true){
int nu1 = nu+i;
int nu2 = nu+j;
if(sw(String.valueOf(nu1))) {
nu=nu1;
break;
}
if(sw(String.valueOf(nu2))) {
nu=nu2;
break;
}
i--;
j++;
}
System.out.println(String.valueOf(nu));
}
}
static boolean sw(String str){
int left = 0;
int right = str.length()-1;
while(left<=right){
if(str.charAt(left)!=str.charAt(right)){
return false;
}
left++;
right--;
}
return true;
}
}
发表于 2021-10-28 01:55:42
回复(0)
简单粗暴的解法，判断是否为奇偶数 奇数则取 前端值 中间值 后端值 偶数则分前后两段进行判断；
判断逻辑：初选对称数字 然后选取上下限度的值进行取值判断
public class Main {
public static long getResultNum(String str) {
long strLong = 0;
try {
strLong = Long.valueOf(str);
} catch (Exception e) {
return strLong;
}
int len = str.length();
if (len < 2) {
return strLong - 1;
}
long result = 0;
/*前半段数据*/
String topVal = null;
/*中间的数据*/
String midVal = null;
/*后半段数据*/
String downVal = null;
if (len % 2 == 0) {
/*偶位数*/
topVal = str.substring(0, len / 2);
downVal = str.substring(len / 2, len);
/*根据topVal获取对称数据*/
String tempStr = reverseVaule(topVal, null);
/*组装成完整对称数据*/
Long resutTemp = Long.valueOf(topVal + tempStr);
/*处理边接问题*/
if (resutTemp > strLong) {
long topLong = Long.valueOf(topVal) - 1;
/*向下取整*/
if (String.valueOf(topLong).length() < len / 2) {
String resultTemp2 = "";
for (int i = 0; i < str.length() - 1; i++) {
resultTemp2 = resultTemp2 + "9";
}
result = Long.valueOf(resultTemp2);
} else {
String reverseVaule = topLong + reverseVaule(topLong + "", null);
result = strLong - (Long.valueOf(reverseVaule)) > (resutTemp - strLong) ? resutTemp : Long.valueOf(reverseVaule);
}
} else {
/*向上取整*/
long topLong = Long.valueOf(topVal) + 1;
if (String.valueOf(topLong).length() > len / 2) {
/*c长度超出则说明超出边界*/
result = Long.valueOf(str) + 2;
} else {
String reverseVaule = topLong + reverseVaule(topLong + "", null);
result = (Long.valueOf(reverseVaule) - strLong) >= (strLong - resutTemp) ? resutTemp : Long.valueOf(reverseVaule);
}
}
} else {
/*奇数*/
topVal = str.substring(0, len / 2);
midVal = String.valueOf(str.charAt((len / 2)));
downVal = str.substring(len / 2 + 1, len);
/*获取对称数据*/
Long temp1 = Long.valueOf(topVal);
Long temp2 = Long.valueOf(topVal + midVal);
Long temp3 = Long.valueOf(downVal);
String resultTemp = topVal + reverseVaule(topVal, midVal);
if (Long.valueOf(resultTemp) >= strLong) {
/*向下取整*/
--temp2;
if (String.valueOf(temp2).length() < (topVal + midVal).length()) {
String reuslt1 = "9";
for (int i = 0; i < String.valueOf(temp3).length(); i++) {
reuslt1 = reuslt1 + "9";
}
result = Long.valueOf(reuslt1);
} else {
String stemp = String.valueOf(temp2);
String topValTemp = stemp.substring(0, len / 2);
String midValTemp = String.valueOf(stemp.charAt((len / 2)));
String resultTemp2 = topVal + reverseVaule(topValTemp, midValTemp);
if (Long.valueOf(resultTemp) == strLong) {
if (String.valueOf(Long.valueOf(topVal + midVal) + 1).length() > (topValTemp + midValTemp).length()) {
result = strLong += 2;
} else {
result = Long.valueOf(resultTemp2);
}
} else {
result = Long.valueOf(resultTemp) - strLong >= strLong - Long.valueOf(resultTemp2) ? Long.valueOf(resultTemp2) : Long.valueOf(resultTemp);
}
}
} else {
/*向上取整*/
++temp2;
if (String.valueOf(temp2).length() > (topVal + midVal).length()) {
result = ++temp2;
} else {
String stemp = String.valueOf(temp2);
String topValTemp = stemp.substring(0, len / 2);
String midValTemp = String.valueOf(stemp.charAt((len / 2)));
String resultTemp2 = topVal + reverseVaule(topValTemp, midValTemp);
result = Long.valueOf(resultTemp2) - strLong >= strLong - Long.valueOf(resultTemp) ? Long.valueOf(resultTemp) : Long.valueOf(resultTemp2);
}
}
}
return result;
}
public static String reverseVaule(String str, String minStr) {
StringBuffer stringBuffer = new StringBuffer();
stringBuffer.append(str);
/*直接反转*/
String reverseTemp = null;
if (minStr == null) {
reverseTemp = stringBuffer.reverse().toString();
} else {
reverseTemp = minStr + stringBuffer.reverse().toString();
}
return reverseTemp;
}
public static void main(String[] args) throws IOException {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
String s = bufferedReader.readLine();
System.out.printf(Main.getResultNum(s)+"");
}
}
编辑于 2021-07-18 01:48:10
回复(0)
思路：取一半长度后面加零，形成一个数。比如数值是145821，得到数145000，来生成145000,146000,144000这三个数的对称数值，145541，146641,144441。再计算最小间隔。
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
String s=sc.next();
long n=Long.parseLong(s);
long m=(long)Math.pow(10, s.length()/2);
long k=n-n%m;
if(k==10)
m=1;
long n1=solve(k+m);
long n2=solve(k);
long n3=solve(k-m);
if(n2==n) {
System.out.println(n2-n3<=n1-n2?n3:n1);
}else if(n2<n){
System.out.println(n-n2<=n1-n?n2:n1);
}else {
System.out.println(n-n3<=n2-n?n3:n2);
}
}
sc.close();
}
private static long solve(long m) {
String s=Long.toString(m);
StringBuilder sb=new StringBuilder();
sb.append(s.substring(0, s.length()/2));
String tmp=sb.reverse().toString();
tmp=s.substring(0, s.length()-s.length()/2)+tmp;
return Long.parseLong(tmp);
}
}
发表于 2021-03-22 22:39:01
回复(0)
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Collections;
public class Main {
/**
* 思路
* 分为数字对称与非对称
*
* 对称要注意取的值要与自己的差值最小
*
* 非对称的数字直接去取对半反转拼接
*
* @param args
* @throws IOException
*/
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
long l = Long.parseLong(s);
StringBuffer sb1 = new StringBuffer();
StringBuffer sb2 = new StringBuffer();
StringBuffer rel = new StringBuffer();
// 取数字中位数
int halfLength = s.length() / 2;
// 前半段放入到sb1中2
for (int i = 0; i < halfLength; i++) {
sb1.append(s.charAt(i));
}
sb2.append(sb1);
// 如果是奇数个数字
if (s.length() % 2 == 1) {
sb2.append(s.charAt(halfLength));
rel.append(sb2);
rel.append(sb1.reverse());
} else {
// 偶数个数字，分情况讨论，分别求出上限、下限、中限 1921
long smallerLeft = (Long.parseLong(sb2.toString()) - 1); //18
long biggerLeft = Long.parseLong(sb2.toString()) + 1; //20
long midLeft = Long.parseLong(sb2.toString()); //19
// 1881
StringBuffer smaller = new StringBuffer();
smaller.append(smallerLeft);
StringBuffer s1 = new StringBuffer(smaller);
smaller.append(s1.reverse());
long l1 = l - Long.parseLong(smaller.toString());
//20021
StringBuffer bigger = new StringBuffer();
bigger.append(biggerLeft);
StringBuffer s2 = new StringBuffer(bigger);
bigger.append(s2.reverse());
long l2 = Long.parseLong(bigger.toString()) - l;
//1991
StringBuffer mid = new StringBuffer();
mid.append(midLeft);
StringBuffer s3 = new StringBuffer(mid);
mid.append(s3.reverse());
long ll2 = l - Long.parseLong(mid.toString());
long ll1 = Long.parseLong(mid.toString()) - l;
long l3 = ll1 >= 0 ? ll1 : ll2;
// 比较差距最小值
Long[] arr = {l1, l2, l3};
Long min = Collections.min(Arrays.asList(arr));
if (l1 == min) {
rel.append(smaller);
} else if (l2 == min) {
rel.append(bigger);
} else {
rel.append(mid);
}
}
System.out.println(rel.toString());
}
}兴业银行笔试时碰到这题，现在写出来了
发表于 2020-11-12 15:04:44
回复(0)
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
import java.util.Queue;
import java.util.Scanner;
import java.util.Stack;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
char[] carr = sc.nextLine().toCharArray();
int mid=(carr.length+1)/2-1;
int i;
for(i=mid+1;i<carr.length;i++) {
carr[i]=carr[carr.length-i-1];
}
for(i=0;i<carr.length;i++)
System.out.print(carr[i]);
}
}
import java.util.Arrays;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
import java.util.Queue;
import java.util.Scanner;
import java.util.Stack;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
char[] carr = sc.nextLine().toCharArray();
int mid=(carr.length+1)/2-1;
int i;
for(i=mid+1;i<carr.length;i++) {
carr[i]=carr[carr.length-i-1];
}
for(i=0;i<carr.length;i++)
System.out.print(carr[i]);
}
}
发表于 2020-06-06 19:45:20
回复(0)
问题信息
通过挑战的用户
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2023-02-08 22:49:07 -
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