uva

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You have devised a new encryption technique which encodes a message by inserting between its characters
randomly generated strings in a clever way. Because of pending patent issues we will not discuss indetail how the strings are generated and inserted into the original message. To validate your method,however, it is necessary to write a program that checks if the message is really encoded in the final string.

Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can remove characters from t such that the concatenation of the remaining characters is s.Input The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII characters separated by whitespace. Input is terminated by EOF. Output For each test case output, if s is subsequence of t.

Sample Input

sequence subsequence
person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter

Sample Output

Yes
No
Yes
No

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import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
	while(in.hasNext()){
		String s1=  in.next();
		String s2 = in.next();
		int index = 0;
		for(int i=0;i < s2.length();i++)
		{
			if(s2.charAt(i) == s1.charAt(index)) index++;
			if(index == s1.length()) break;
		}
		if(index == s1.length()) System.out.println("Yes");
		else System.out.println("No");
	}
	}
}

小结

if(index == s1.length()) break;

这一行代码最值得注意。没有这行代码，则会出现下面这种情况

所以清注意思考这个地方为什么会报错。

其实代码出错在这个 ，看我的测试

此时换行符相等了，index增加到了5，导致判断会出错，因为我们看到原本有5个字符串，应该index只到4，这里却由于换行符的干扰导致错误，所以应该index一满就break；这样也节约了时间。