考拉有n个字符串字符串，任意两个字符串长度都是不同的。考拉最近学习到有两种字符串的排序方法： 1.根据字符串的字典序排序。例如：
"car" < "carriage" < "cats" <
"doggies < "koala"
2.根据字符串的长度排序。例如：
"car" < "cats" < "koala" <
"doggies" < "carriage"
考拉想知道自己的这些字符串排列顺序是否满足这两种排序方法，考拉要忙着吃树叶，所以需要你来帮忙验证。
[编程题]两种排序方法
- 热度指数：32712 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 32M，其他语言64M
- 算法知识视频讲解
输入描述:
输入第一行为字符串个数n(n ≤ 100) 接下来的n行,每行一个字符串,字符串长度均小于100，均由小写字母组成
输出描述:
如果这些字符串是根据字典序排列而不是根据长度排列输出"lexicographically",
如果根据长度排列而不是字典序排列输出"lengths",
如果两种方式都符合输出"both"，否则输出"none"
示例1
输入
3 a aa bbb
输出
both
import java.util.*;
public class Main{
public static boolean iszd(String[] str){
for(int i= 0; i<str.length-1; i++){
if(str[i].compareTo(str[i+1]) > 0){
return false;
}
}
return true;
}
//字符串长度判断
public static boolean iscd(String[] str){
for(int i= 0; i<str.length-1; i++){
if(str[i].length() > str[i+1].length()){
return false;
}
}
return true;
}
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int n = Integer.valueOf(sc.nextLine());
String[] str = new String[n];
for(int i = 0;i < n;i++){
str[i] = sc.nextLine();
}
// boolean zd = iszd(arr);
// boolean cd = iscd(arr);
if(iszd(str) && iscd(str)){
System.out.println("both");
}else if(iszd(str)){
System.out.println("lexicographically");
}else if(iscd(str)){
System.out.println("lengths");
}else{
System.out.println("none");
}
}
}
发表于 2022-03-30 17:53:17
回复(0)
public class Main{
public static boolean isLengths(String[] str){//判断是不是按长度排列
for(int i=0;i< str.length-1;i++){
if(str[i].length()>str[i+1].length()){
return false;
}
}
return true;
}
//判断两个字符串是不是字典序排列——str1是不是在字典中是在str2前面
public static boolean isLex(String str1,String str2){
//字典序的话字母相同也是要分长度的：比如car<carriage
int l=min(str1.length(),str2.length());//找出str1和str2中较小长度（不能找较大长度，否则小的那个会越界）
for(int i=0;i<l;i++){
if(str1.charAt(i)<str2.charAt(i)){
return true;
}
if(str1.charAt(i)>str2.charAt(i)){
return false;
}
}
//走到这里没有return 说明前面一部分字母都是相同的
//但是car在字典中不能排在carriage后——也就是str1=="carriage",str2=="car"不能直接返回true
if(str1.charAt(l-1)==str2.charAt(l-1)){
//不能直接if(str1.length()>str2.length())
//有可能出现carriage 和 cats这种情况，本来是字典序没问题，
// 但是前者长度大于后者会进入里面的if，你直接return false就错了，所以外层套一个if，判断末尾结束字符是否一样
if(str1.length()>str2.length()){
return false;
}
}
return true;
}
public static boolean isLexicographically(String[] str){//判断是不是字典序排列
int k=0;//标记对比次数，如果要return true,需要全部对比完
// 比如字符串组str中有3个，你要对比1和2,2和3，共对比2次。如果str中有n个，要对比n-1次
for(int i=0;i<str.length-1;i++){
if(isLex(str[i],str[i+1])){
k++;
if(k==str.length-1){
return true;
}
}else{
break;//进入else说明上面isLex返回的是false，也就是：不是字典排序
}
}
return false;
}
public static void main(String[]args){
Scanner scanner=new Scanner(System.in);
int n=scanner.nextInt();
scanner.nextLine();//把n产生的回车回收一下
String[] arr=new String[n];
int x=0;//标记是不是按长度排列，不是——x=0，是——x=1
int y=0;//标记是不是按字典序排列，不是——y=0，是——y=1
for(int i=0;i<n;i++){
arr[i]=scanner.nextLine();
}
if(isLengths(arr)){
x=1;
}
if(isLexicographically(arr)){
y=1;
}
if(y==1&&x==0){//根据字典序排列,但不是根据长度排列
System.out.println("lexicographically");
}
if(y==0&&x==1){//根据长度排列,但不是根据字典序排列
System.out.println("lengths");
}if(y==1&&x==1){//都是
System.out.println("both");
}if(y==0&&x==0){//都不是
System.out.println("none");
}
}
} 写了一个多小时终于全部通过了
发表于 2022-03-28 21:35:29
回复(1)
import java.util.*;
public class Main {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
int n = sc.nextInt();
String[] original = new String[n];
PriorityQueue<String> strQ = new PriorityQueue<>(new Comparator<String>(){
public int compare(String s1, String s2){
return s1.compareTo(s2);
}
});
PriorityQueue<String> lenQ = new PriorityQueue<>(new Comparator<String>(){
public int compare(String s1, String s2){
return s1.length() - s2.length();
}
});
for(int i = 0; i < n; i++){
String cur = sc.next();
original[i] = cur;
strQ.add(cur);
lenQ.add(cur);
}
boolean strFlag = true;
boolean lenFlag = true;
int index = 0;
while(!strQ.isEmpty()){
if(original[index] != strQ.poll()){
strFlag = false;
}
if(original[index] != lenQ.poll()){
lenFlag = false;
}
index++;
}
if(strFlag && lenFlag){
System.out.println("both");
}else if(strFlag){
System.out.println("lexicographically");
}else if(lenFlag){
System.out.println("lengths");
}else{
System.out.println("none");
}
}
sc.close();
}
}
发表于 2020-08-17 15:11:24
回复(0)
import java.util.Scanner;
public class Main {
static boolean ifZidian = true;
static boolean ifLength = true;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = Integer.valueOf(sc.nextLine());
String[] s = new String[n];
for(int i=0;i<n;i++) {
s[i] = sc.nextLine();
}
for(int i=0;i<n-1;i++) {
if(ifZidian==false && ifLength==false) break;
if(s[i].compareTo(s[i+1])>0) ifZidian=false;
if(s[i].length()-s[i+1].length()>0) ifLength = false;
}
if(ifZidian==true && ifLength==false) System.out.println("lexicographically");
else if(ifZidian==false && ifLength==true) System.out.println("lengths");
else if(ifZidian==true && ifLength==true) System.out.println("both");
else System.out.println("none");
}
}
public class Main {
static boolean ifZidian = true;
static boolean ifLength = true;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = Integer.valueOf(sc.nextLine());
String[] s = new String[n];
for(int i=0;i<n;i++) {
s[i] = sc.nextLine();
}
for(int i=0;i<n-1;i++) {
if(ifZidian==false && ifLength==false) break;
if(s[i].compareTo(s[i+1])>0) ifZidian=false;
if(s[i].length()-s[i+1].length()>0) ifLength = false;
}
if(ifZidian==true && ifLength==false) System.out.println("lexicographically");
else if(ifZidian==false && ifLength==true) System.out.println("lengths");
else if(ifZidian==true && ifLength==true) System.out.println("both");
else System.out.println("none");
}
}
发表于 2018-07-21 17:13:06
回复(0)
import java.util.Scanner;
public class SortString {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int length=sc.nextInt();
String [] str=new String[length];
for(int i=0;i<length;i++){
str[i]=sc.next();
}
sortString(str,length);
}
private static void sortString(String[] str, int length) {
boolean a=true;
int len[] =new int[length];
for(int i=0;i<length;i++){
len[i]=str[i].length();
}
//确定是否为长度递增
for(int i=0;i<length-1;i++){
if(len[i+1]-len[i]<0){
a=false;
}
}
//去顶是否为字典增
boolean b=true;
for(int i=0;i<length-1;i++){
//String 方法 compareTo 得int型变量 大于0 书名是字典序增的
int temp=str[i].compareTo(str[i+1]);
if(temp>0){b=false;}
}
if(a&&b){
System.out.println("both");
}
if(a&&(!b)){
System.out.println("lengths");
}
if((!a)&&b){
System.out.println("lexicographically");
}
if((!a)&&(!b)){
System.out.println("none");
}
}
}
public class SortString {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int length=sc.nextInt();
String [] str=new String[length];
for(int i=0;i<length;i++){
str[i]=sc.next();
}
sortString(str,length);
}
private static void sortString(String[] str, int length) {
boolean a=true;
int len[] =new int[length];
for(int i=0;i<length;i++){
len[i]=str[i].length();
}
//确定是否为长度递增
for(int i=0;i<length-1;i++){
if(len[i+1]-len[i]<0){
a=false;
}
}
//去顶是否为字典增
boolean b=true;
for(int i=0;i<length-1;i++){
//String 方法 compareTo 得int型变量 大于0 书名是字典序增的
int temp=str[i].compareTo(str[i+1]);
if(temp>0){b=false;}
}
if(a&&b){
System.out.println("both");
}
if(a&&(!b)){
System.out.println("lengths");
}
if((!a)&&b){
System.out.println("lexicographically");
}
if((!a)&&(!b)){
System.out.println("none");
}
}
}
发表于 2018-07-09 11:04:45
回复(0)
Java中的String默认按照字典序进行排序，字典序相同再按长度排序。
要按长度排序直接实现一个比较器即可，这里直接用lambda表达式
import java.util.Arrays;
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String line = scanner.nextLine();
int n = Integer.parseInt(line);
String []s = new String[n];
for(int i = 0; i < n;i ++) {
line = scanner.nextLine();
s[i] = line;
}
if (lexicographically(s) && lengths(s)) {
System.out.println("both");
}
else if (lexicographically(s)) {
System.out.println("lexicographically");
}
else if (lengths(s)) {
System.out.println("lengths");
}
else {
System.out.println("none");
}
}
public static boolean lexicographically(String []s) {
String[] sort = Arrays.copyOf(s, s.length);
Arrays.sort(sort);
for (int i = 0;i < s.length;i ++) {
if (!s[i].equals(sort[i])) {
return false;
}
}
return true;
}
public static boolean lengths(String []s) {
String[] sort = Arrays.copyOf(s, s.length);
Arrays.sort(sort, (a,b) -> a.length() - b.length());
for (int i = 0;i < s.length;i ++) {
if (!s[i].equals(sort[i])) {
return false;
}
}
return true;
}
}
发表于 2018-05-21 15:54:22
回复(0)
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
int n = sc.nextInt();
boolean[] flags = {true, true}; // {lexicographically, lengths}
String[] arr = new String[n];
for (int i = 0; i < n; i++) arr[i] = sc.next();
for (int i = 0, j = 1; i < n - 1 && j < n; i++, j++) {
String s1 = arr[i];
String s2 = arr[j];
if (s1.compareTo(s2) > 0) flags[0] = false;
if (s1.length() > s2.length()) flags[1] = false;
if (!flags[0] && !flags[1]) break;
}
if (flags[0] && flags[1]) System.out.println("both");
else if (flags[0] && !flags[1]) System.out.println("lexicographically");
else if (!flags[0] && flags[1]) System.out.println("lengths");
else System.out.println("none");
}
}
}
发表于 2018-04-10 03:39:06
回复(0)
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scanner = new Scanner(System.in);
int n = Integer.parseInt(scanner.nextLine());
String str[] = new String[n];
boolean lexicographically = true;
boolean lengths = true;
for (int i = 0; i < n; i++) {
str[i] = scanner.nextLine();
if (i != 0) {
lexicographically = lexicographically && isGreater(str[i - 1], str[i]);
lengths = lengths && (str[i].length() > str[i - 1].length());
}
}
if (lexicographically && lengths)
System.out.println("both");
else if (lexicographically)
System.out.println("lexicographically");
else if (lengths)
System.out.println("lengths");
else
System.out.println("none");
}
public static boolean isGreater(String a, String b) {
int len = Math.min(a.length(), b.length());
boolean flag = true;
for (int i = 0; i < len; i++) {
if (a.charAt(i) > b.charAt(i))
return false;
if (a.charAt(i) < b.charAt(i))
return true;
}
if (a.length() > b.length())
return false;
return true;
}
// TODO Auto-generated method stub
Scanner scanner = new Scanner(System.in);
int n = Integer.parseInt(scanner.nextLine());
String str[] = new String[n];
boolean lexicographically = true;
boolean lengths = true;
for (int i = 0; i < n; i++) {
str[i] = scanner.nextLine();
if (i != 0) {
lexicographically = lexicographically && isGreater(str[i - 1], str[i]);
lengths = lengths && (str[i].length() > str[i - 1].length());
}
}
if (lexicographically && lengths)
System.out.println("both");
else if (lexicographically)
System.out.println("lexicographically");
else if (lengths)
System.out.println("lengths");
else
System.out.println("none");
}
public static boolean isGreater(String a, String b) {
int len = Math.min(a.length(), b.length());
boolean flag = true;
for (int i = 0; i < len; i++) {
if (a.charAt(i) > b.charAt(i))
return false;
if (a.charAt(i) < b.charAt(i))
return true;
}
if (a.length() > b.length())
return false;
return true;
}
发表于 2018-03-07 15:04:06
回复(0)
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int count = in.nextInt(); in.nextLine();
List<String> list = new ArrayList<>();
for (int i = 0; i < count; i++)
list.add(in.nextLine());
List<String> listCopy = new ArrayList<>(list);
Collections.sort(listCopy);
boolean isSorted1 = false, isSorted2 = true;
if (listCopy.equals(list)) isSorted1 = true;
for (int i = 0; i < count - 1; i++) {
if (list.get(i).length() > list.get(i+1).length())
isSorted2 = false;
}
if (isSorted1 && isSorted2) System.out.println("both");
else if (isSorted1 && !isSorted2) System.out.println("lexicographically");
else if (!isSorted1 && isSorted2) System.out.println("lengths");
else System.out.println("none");
}
}
---------------------------------------------------------------------------
//参考高票答案
import java.util.*;
import static java.lang.System.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int count = in.nextInt();
String[] strings = new String[count];
for (int i = 0; i < count; i++)
strings[i] = in.next();
boolean isLexicographical = isLexicographical(strings);
boolean isSortedByLen = isSortByLen(strings);
if (isLexicographical && isSortedByLen) out.println("both");
else if (isLexicographical) out.println("lexicographically");
else if (isSortedByLen) out.println("lengths");
else out.println("none");
}
private static boolean isSortByLen(String[] strings) {
boolean res = true;
for (int i = 0; i < strings.length - 1; i++)
if (strings[i].length() > strings[i + 1].length())
res = false;
return res;
}
private static boolean isLexicographical(String[] strings) {
boolean res = true;
for (int i = 0; i < strings.length - 1; i++)
if (strings[i].compareTo(strings[i + 1]) > 0)
res = false;
return res;
}
}
编辑于 2017-12-18 19:57:36
回复(0)
import java.util.LinkedList; import java.util.Scanner; public class Main { public static void main(String args[]) { Scanner scanner = new Scanner(System.in); LinkedList<String> strings = new LinkedList<>(); while (scanner.hasNext()) { int n = Integer.parseInt(scanner.nextLine()); for (int i = 0; i < n; i++) { strings.add(i, scanner.nextLine()); } boolean A = ifLengths(strings); boolean B = ifLexicographically(strings); if (A && B) { System.out.println("both"); } else if (A) { System.out.println("lengths"); } else if (B) { System.out.println("lexicographically"); } else { System.out.println("none"); } } } private static boolean ifLexicographically (LinkedList<String> strings) { int length = strings.size(); int i; for (i = 0; i < length - 1; i++) { if (strings.get(i + 1).compareTo(strings.get(i)) < 0) { break; } } return (i == length - 1); } private static boolean ifLengths (LinkedList<String> strings) { int length = strings.size(); int i; for (i = 0; i < length - 1; i++) { if (strings.get(i + 1).length() < strings.get(i).length()) { break; } } return (i == length - 1); } }这个题其实蛮简单的，因为字符串的compare（）函数直接就可以解决两个字符串按字典序的先后顺序，然后length（）又可以直接确定按照长度的先后顺序，所以这道题是蛮简单的
发表于 2017-11-23 19:42:20
回复(0)
两个字符串使用字典排序进行判断使用：compareTo()；
import java.util.Scanner;
public class StringUtil{
//两种排序方法
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while(in.hasNext()){
int n = Integer.valueOf(in.nextLine());
String str[] = new String[n+1];
for(int i=1; i<=n; i++){
str[i] = in.nextLine();
}
boolean zd = iszd(str);
boolean cd = iscd(str);
if(zd && cd)
System.out.println("both");
else if(zd)
System.out.println("lexicographically");
else if(cd)
System.out.println("lengths");
else
System.out.println("none");
}
}
//字典判断
public static boolean iszd(String[] str){
for(int i=1; i<str.length-1; i++){
if(str[i].compareTo(str[i+1]) >= 0)
return false;
}
return true;
}
//字符串长度判断
public static boolean iscd(String[] str){
for(int i=1; i<str.length-1; i++){
if(str[i].length() >= str[i+1].length())
return false;
}
return true;
}
}
发表于 2017-09-07 01:25:33
回复(0)
import java.util.Scanner;
public class Main {
public static int compare(char[] c1, char[] c2) {
int i = 0;
for (i = 0; i < c1.length && i < c2.length; i++) {
if (c1[i] != c2[i])
break;
}
if (i < c1.length && i < c2.length) {
return c2[i] - c1[i];
}
return c2.length - c1.length;
}
public static void main(String[] args) {
try (Scanner scanner = new Scanner(System.in)) {
A: while (scanner.hasNext()) {
int n = scanner.nextInt();
if (n <= 1)
System.out.println("both");
char[] c1 = scanner.nextLine().toCharArray();
boolean isLenSort = true, isDirSort = true;
char[] last = c1;
for (int i = 1; i <= n; i++) {
char[] c = scanner.nextLine().toCharArray();
if (isLenSort && (c.length < last.length))
isLenSort = false;
if (isDirSort && compare(c, last) > 0)
isDirSort = false;
// if(!isLenSort&&!isDirSort)
// continue A;
last = c;
}
if (isLenSort && isDirSort)
System.out.println("both");
else if (isLenSort)
System.out.println("lengths");
else if (isDirSort)
System.out.println("lexicographically");
else
System.out.println("none");
}
}
}
}
k eng 坑->只能由小到大
发表于 2017-08-16 15:33:18
回复(0)
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int n = Integer.parseInt(sc.nextLine());
ArrayList<String> s = new ArrayList<String>();
for(int i = 0; i < n; i++)
s.add(sc.nextLine());
if(isDicSort(s) && (!isLagIng(s)))
System.out.print("lexicographically");
else if((!isDicSort(s)) && isLagIng(s))
System.out.print("lengths");
else if(isDicSort(s) && isLagIng(s))
System.out.print("both");
else if(!isDicSort(s) && !isLagIng(s))
System.out.print("none");
}
public static boolean isDicSort(ArrayList<String> s){
ArrayList<String> s1 = (ArrayList<String>)s.clone();
Collections.sort(s1);
for(int i = 0; i < s.size(); i++){
if(!s.get(i).equals(s1.get(i)))
return false;
}
return true;
}
public static boolean isLagIng(ArrayList<String> s){
for(int i = 1; i < s.size(); i++){
if(s.get(i).length() <= s.get(i-1).length())
return false;
}
return true;
}
}
发表于 2017-08-11 19:34:50
回复(0)
import java.util.*;
public class Main
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int n=0;
ArrayList<String> list = null;
if(sc.hasNext())
{
n = sc.nextInt();
sc.nextLine();//行末多个换行。需要跳到第二行
list = new ArrayList<>(n);
for(int i =0;i<n;i++)
{
list.add(sc.nextLine().trim());
}
}
boolean isLex = true;
boolean isLen = true;
int len = list.get(0).length();
for(int i = 0;i<list.size()-1;i++)
{
if(list.get(i).compareTo(list.get(i+1))>0)
isLex =false;
if(list.get(i).length()>list.get(i+1).length())
isLen=false;
}
if(isLex)
{
if(isLen)
System.out.println("both");
else
System.out.println("lexicographically");
}
else
{
if(isLen)
System.out.println("lengths");
else
System.out.println("none");
}
sc.close();
}
}
发表于 2017-08-10 15:35:03
回复(0)
问题信息
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2023-03-14 02:46:18
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