实现简易字符串压缩算法:由字母a-z或者A-Z组成,将其中连续出现2次以上(含2次)的字母转换为字母和出现次数,

@Test
public void test1(){
   String content1 = "AAAAAAAAAAAAAAAAAAAAAAAAttBffgfaaddddddsCDaaaBBBBdddfdsgggggg";
   String result = yasuo(content1);
    System.out.println(result);
}

public String yasuo(String content){
    content = content +" ";
    StringBuilder builder = new StringBuilder();
    int time = 1;
    for(int i=0;i<content.length()-1;i++){
        String s = content.charAt(i)+"";
        String next = content.charAt(i+1)+"";
        if(s.matches("\"^[0-9]*$")){
            builder.append(s);
        }
        if(s.equals(next)){
            time+=1;
        }else{
            if(time!=1){
                builder.append(time+s);
                time = 1;
                continue;
            }
            builder.append(s);
        }
    }
    return builder.toString();
}
打印结果:
 

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转载自blog.csdn.net/zhanglei082319/article/details/79194451
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