题解 | #单词倒排#
单词倒排
https://www.nowcoder.com/practice/81544a4989df4109b33c2d65037c5836
我本地的JDK是16 下面这个正则表达式 分词是没有问题的
// [I, am, a, student]
System.out.println(Arrays.toString("I am a student.".split("[^[a-zA-Z]]+")));
但是牛客的JDK是 1.8 分词是有问题的// [, , , , .]
System.out.println(Arrays.toString("I am a student.".split("[^[a-zA-Z]]+")));
改成下面这样就行 import java.util.*;
/**
* @author hll[yellowdradra@foxmail.com]
* @date 2022-09-30 15:53
**/
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
Arrays.stream(in.nextLine()
// 分割字符串得到一个单词数组
.split("[^a-zA-Z]+"))
// 反转数组
.sorted((x, y) -> -1)
// 合并数组
.reduce((x, y) -> x + " " + y)
// 打印数组
.map(System.out::printf);
}
}
下面是双指针解法
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String str = in.nextLine();
StringBuilder sb = new StringBuilder();
int len = str.length(), l = 0, r = 0, i = len - 1;
while (i >= 0) {
while (i >= 0) {
if (isLetter(str.charAt(i))) {
r = i + 1;
break;
}
i--;
}
while (i >= 0) {
if (!isLetter(str.charAt(i))) {
l = i + 1;
i--;
break;
}
if (isLetter(str.charAt(i)) && i == 0) {
l = 0;
i--;
break;
}
i--;
}
sb.append(str, l, r).append(" ");
}
System.out.println(sb);
}
public static boolean isLetter(char c) {
return ('a' <= c && c <= 'z') || ('A' <= c && c <= 'Z');
}
}
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