1. 问题描述:
例如:从控制台输入here are you,输出you are here
思路是使用Java提供翻转字符串的方法,先翻转整个字符串,然后使用正则表达式对翻转后的字符串进行分割,这里使用空格分割,然后把分割后的结果存储到一个String数组中,之后再for循环中遍历然后再把每个单词翻转回来,具体的代码如下:
import java.util.Scanner;
public class Main{
//测试用例: here are you
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
solve(str);
}private static void solve(String str) {
String s = reverseString(str);// 匹配多个空格的正则表达式regex = "\\s+"
String words[] = s.split("\\s+");
StringBuilder sb = new StringBuilder();
for(int i = 0; i < words.length; i++){//加入翻转后的单词
sb.append(reverseString(words[i])+" ");
}
System.out.println(sb.toString());
}private static String reverseString(String str) {
StringBuilder sb = new StringBuilder(str);
return sb.reverse().toString();
}
}
也可以这样写:先分割最开始的字符串串然后遍历最开始的字符串遇到空格加入空格,具体的代码如下:
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
solve(str);
}private static void solve(String str) {
String words[] = str.split("\\s+");
StringBuilder sb = new StringBuilder();
int count = words.length - 1;
for(int i = str.length() - 1; i>=0; i--){
if(str.charAt(i) == ' '){
sb.append(words[count--] + " ");
}
int j = i;
while(str.charAt(j--) == ' ');
i = j + 2;
}
sb.append(words[0]);
System.out.println(sb.toString());
}
}
有时候需要对其中单词之间的多个空格也要进行输出,此时就需要在循环中记录空格的数量然后加入到字符串中,具体的代码如下:
import java.util.Scanner;
public class Main{
//测试用例: here are you
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
solve(str);
}private static void solve(String str) {
String words[] = str.split("\\s+");
StringBuilder sb = new StringBuilder();
int count = words.length - 1;
for(int i = str.length() - 1; i >= 0 ; i--){
if(str.charAt(i) == ' '){
int j = i;
sb.append(words[count--]);
while(str.charAt(j--) == ' '){
sb.append(str.charAt(j + 1));
}
i = j + 2;
}
}
sb.append(words[0]);
System.out.println(sb.toString());
}
}