Java 字符串压缩打印错误的字符串。我在这里缺少什么？

Question

我有一个字符串“aaabbaaaavvvd”，它应该被压缩为a3b2a4v3d但是当我运行代码时它会打印出 b3a2v4d3

奇怪的是，它以 b 开头，而不是 a

public class compression {
    public String compress(String str){
        char chararr[] = str.toCharArray();
        StringBuilder sb = new StringBuilder();
        int count=1;
        char previous = chararr[0];
        for(int i=1; i<chararr.length; i++) {
             char current = chararr[i]; 
             if(current == previous){
                 count++;
             } else {
                 sb.append(current).append(count);
                 count = 1;
             }
             previous = current;
        }
        System.out.println(sb.toString());
        return sb.toString();
    }

    public static void main(String args[]){
        compression test = new compression();
        String str = "aaabbaaaavvvd";
        test.compress(str);
    }
}

Answer 1

线路错误:

sb.append(current).append(count);

考虑一下 if 语句

if(current == previous){
    count++;
} else {
    sb.append(current).append(count);
    count = 1;
}

当您的代码逐步执行每个字符时，仅当当前字符不等于最后一个字符(即当前字符是新的字符序列)。

逐一进行:

Index 1: previous == 'a', current == 'a', if-statement: true - increments
Index 2: previous == 'a', current == 'a', if-statement: true - increments
Index 3: previous == 'a', current == 'b', if-statement: false - prints and reset

请注意，在第三个索引上，current 是 b 而不是所需的字符 a，输出 b3与a3相反。

用以前的替换附加链来修复:

if(current == previous){
    count++;
} else {
    sb.append(previous).append(count);
    count = 1;
}