题解 | #万万没想到之聪明的编辑#

主要思路,就是分两次处理, 利用数组标记是否 被 删除 过, 最开始想一次处理完, 发现不行.

import java.util.Scanner;

/**
 * @Author Fizz Pu
 * @Date 2021/6/19 下午4:10
 * @Version 1.0
 * 失之毫厘，缪之千里！
 */
public class Main1 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int n = scanner.nextInt();
        for(int q = 0; q < n; q ++) {
            String data = scanner.next();

            char[] arrays = data.toCharArray();
            int len = arrays.length;

            if (len < 3) {
                System.out.println(data);
                continue;
            }

            // 第一次处理 AAA 型
            int[] delete = new int[len]; // 1为删除
            int i = 0, j = 1, k = 2;
            int count =  len; // 字符串的个数
            while (k < len) {
                if (arrays[i] == arrays[j] && arrays[i] == arrays[k]) {
                    delete[k] = 1;
                    count--;
                }

                i ++;
                j ++;
                k ++;
            }

            //  生成新的数组
            // 在第一次的基础上处理AABB型
            char[] newStr  =  new  char[count];
            i = 0;
            for (int m = 0; m < len; m++) {
                if (delete[m] == 0) {
                    newStr[i++] = arrays[m];
                }
                delete[m] = 0;
            }

            i = 0;
            j = 1;
            k = 2;
            while (k < count - 1) {
                if (newStr[i] == newStr[j] && newStr[k] == newStr[k+1]) {
                    delete[k] = 1;
                    i = k + 1;
                    j = i + 1;
                    k = j + 1;
                }
                i++;
                j++;
                k++;
            }

            for (int m = 0; m < count; m++) {
                if (delete[m] == 0) {
                    System.out.print(newStr[m]);
                }
            }
            System.out.println();
        }
    }
}