Time Limit: 2000/1000 MS (java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 29094 Accepted Submission(s): 11433
Ignatius likes to write Words in reverse way. Given a single line of text which is written by Ignatius, you should reverse all the words and then output them.
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. Each test case contains a single line with several words. There will be at most 1000 characters in a line.
For each test case, you should output the text which is processed.
3 olleh !dlrow m’I morf .udh I ekil .mca
hello world! I’m from hdu. I like acm.
题解：字符串反串，可以运用栈原理。
import java.util.Scanner;public class Main { public static String revert(String s){ String res = ""; char[] s1 = s.toCharArray(); for(int i=s1.length-1;i>=0;i--){ res+=s1[i]; }// System.out.println(res); return res; } public static void main(String[] args) { Scanner in = new Scanner(System.in); int N = in.nextInt(); in.nextLine(); while(N-- > 0){ String str = in.nextLine();// System.out.println(str.length()); //字符串处理 String[]strs = str.split(" "); String res = "";// System.out.println("length"+strs.length); for(int i=0;i<strs.length;i++){ if(i==strs.length-1) res += revert(strs[i]); else res += revert(strs[i]) + " "; } if(str.endsWith(" ")) res += " ";//Java代码需要考虑这种情况，不然提交PE。 //PE后反思了一下发现还是自己写的方法靠谱一些，处理出的东西不是杂七杂八的 System.out.println(res); } }}这是我自己没用Java库提供的方法处理字符串，居然简洁的不得了，只有20多行代码，啊哈哈，而且一次AC
import java.util.Scanner;public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int N = in.nextInt(); in.nextLine(); while(N-- > 0){ String str = in.nextLine(); //字符串处理 String res = ""; String res1 = ""; for(int i=0;i<str.length();i++){ if(str.charAt(i) == ' '){ res += res1; res += str.charAt(i); res1 = ""; }else{ res1 = str.charAt(i) + res1; } } res += res1; System.out.println(res); } }}新闻热点
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