我的PAT-BASIC代碼倉:https://github.com/617076674/PAT-BASIC
原題連結:https://pintia.cn/problem-sets/994805260223102976/problems/994805314941992960
題目描述:
![](https://img.laitimes.com/img/__Qf2AjLwojIjJCLyojI0JCLiAzNvwVZ2x2bzNXak9CX90TQNNkRrFlQKBTSvwFbslmZvwFMwQzLcVmepNHdu9mZvwFVywUNMZTY18CX052bm9CXykEVPhXTq1EeRpHW4Z0MMBjVtJWd0ckW65UbM5WOHJWa5kHT20ESjBjUIF2LcRHelR3LcJzLctmch1mclRXY39DO3UDO1MDN5EDMyATM4EDMy8CX0Vmbu4GZzNmLn9Gbi1yZtl2Lc9CX6MHc0RHaiojIsJye.jpg)
知識點:字元串讀取
思路:讀入所有字元串進一個集合,再倒序輸出
對于C++而言,可以用getchar() == '\n'來判斷是否已經全部輸入完畢。
而對于Java而言,可以讀取整行資料再用split函數分段得到每一個字元串。
時間複雜度和空間複雜度均是O(n),其中n為輸入字元串個數。
C++代碼:
#include<iostream>
#include<vector>
#include<string>
using namespace std;
int main() {
string temp;
vector<string> strings;
while (true) {
cin >> temp;
strings.push_back(temp);
if (getchar() == '\n') {
break;
}
}
for (int i = strings.size() - 1; i >= 0; i--) {
cout << strings[i];
if (i != 0) {
cout << " ";
}
}
}
C++解題報告:
JAVA代碼:
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String input = scanner.nextLine();
String[] strings = input.split(" ");
for (int i = strings.length - 1; i >= 0; i--) {
System.out.print(strings[i]);
if(i != 0){
System.out.print(" ");
}
}
}
}
JAVA解題報告: