杭电ACM 1020 Encoding java解析
最后更新于：2022-04-01 09:48:39
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# Encoding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34681 Accepted Submission(s): 15377
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Problem Description
Given a string containing only 'A' - 'Z', we could encode it using the following method:
1. Each sub-string containing k same characters should be encoded to "kX" where "X" is the only character in this sub-string.
2. If the length of the sub-string is 1, '1' should be ignored.
Input
The first line contains an integer N (1 <= N <= 100) which indicates the number of test cases. The next N lines contain N strings. Each string consists of only 'A' - 'Z' and the length is less than 10000.
Output
For each test case, output the encoded string in a line.
Sample Input
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2
ABC
ABBCCC
~~~
Sample Output
ABC
A2B3C解题思路整理：对字符串进行遍历，如果当前的字符与前面的字符相同的话那么我加1记录其个数，如果不相同的时候，排除只有1个的情况，只有在字符重复个数大于1的时候那么认为是有多个在一起，这是将字符串拼接组成新的串继续下一步操作。
~~~
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int count = scanner.nextInt();
int i = 0, countNum = 0, j = 0, j2 = 0;
String str ;
String str2 = "";
char sCopy = 0;
List<String> list = new ArrayList<String>();
while(i < count)
{
str = scanner.next();
sCopy = 0;
j = 0;
str2 = "";
countNum = 0;
while(j < str.length())
{
if(sCopy != str.charAt(j))
{
if(countNum > 1)
{
str2 = str2 + countNum + sCopy;
}
else if(j != 0)
str2 = str2 + sCopy;
countNum = 0;
sCopy = str.charAt(j);
}
if(sCopy == str.charAt(j))
{
countNum++;
}
j++;
if(j == str.length())
{
if(countNum > 1)
{
str2 = str2 + countNum + sCopy;
}
else if(j != 0)
str2 = str2 + sCopy;
list.add(str2);
}
}
i++;
}
for(i = 0; i < count; i++)
{
System.out.println(list.get(i));
}
}
}
~~~