HDU---1020(基本String操作)

最新推荐文章于 2021-08-13 19:04:07 发布

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#字符串处理 #HDU

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这篇博客介绍了HDU 1020问题的解决方案，涉及字符串编码方法：连续相同字符的子串会被编码为'kX'形式，其中X是字符，k是其出现次数。题目给出输入输出样例，并分析了编程实现时可能的错误点，包括未针对连续字符整合和误处理末尾换行。

Encoding

Problem Description

Given a string containing only ‘A’ - ‘Z’, we could encode it using the following method:

Each sub-string containing k same characters should be encoded to “kX” where “X” is the only character in this sub-string.
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

2
ABC
ABBCCC

Sample Output

ABC
A2B3C

分析:

将每个字母所连续的次数k, 整合成”k字母”的形式输出

错误点1: 没有仔细读题, 将整个字符串中所有同种字母整合成k的情况
错误点2: 不需要判断最后一行没有换行, 如果你不幸这么做了,就会的到一个格式错误(坑爹!)

AC代码:

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int num = sc.nextInt();
        String str = null;
        int index, count;
        while(num-- > 0) {
            str = sc.next() + "#";
            index = 0;
            while(str.charAt(index) != '#') {
                count = 1;
                while(str.charAt(index) == str.charAt(index + 1)) {
                    count++;
                    index++;
                }
                if(count > 1) {
                    System.out.print("" + count + str.charAt(index));
                } else {
                    System.out.print(str.charAt(index));
                }
                index++;
            }
            System.out.println();
        }
        sc.close();
    }
}