java二进制数据的切割和数量计算

Question

这段对二进制数据切割的代码是对一整串二进制数据按规则切割。连续1遇0切割，连续0遇1切割，带X的用规则切割。可以加个计算出切割后以1开头的字符串和以0开头的字符串的数量的代码吗。

import java.io.;
public class aa {
/*
* 在源数据中将指定字符替换成最后的数字
* @param sourceData
* @param repalceChar
* @param index
* @return
*/
public static String replaceDataFromTail(String sourceData, String repalceChar,int index) {
if (!sourceData.substring(index, index + 1).equals(repalceChar)) {
return sourceData.replaceFirst(repalceChar, sourceData.substring(index, index+1));
}
if (sourceData.substring(index, index + 1).equals(repalceChar)&& (index== sourceData.length()-1)) {
return sourceData.replaceFirst(repalceChar, sourceData.substring(index, index+1));
}
return replaceDataFromTail(sourceData, repalceChar,index+1);
}

/**
 * 在源数据中用前面的数字替换指定字符
 * @param sourceData
 * @param repalceChar
 * @param index
 * @return
 */
public static String replaceDataFromHead(String sourceData, String repalceChar,int index) {
    if (index> sourceData.length()-1) {
        return  sourceData;
    }
    if (!sourceData.substring(index,index+1).equals(repalceChar)) {
        return sourceData;
    }
    String prevOneChar= "";
    String prevTwoChar= "";
    String prevThreeChar= "";
    switch (index)
    {
        case 0:
            break;
        case 1:
            prevOneChar = sourceData.substring(index - 1, index);
            sourceData = sourceData.replaceFirst(repalceChar, prevOneChar);
            sourceData = replaceDataFromHead(sourceData, repalceChar, index + 1);

            break;
        case 2:
            prevOneChar = sourceData.substring(index - 1, index);
            prevTwoChar= sourceData.substring(index-2, index-1);
            if(prevOneChar.equals(prevTwoChar)) {
                sourceData = sourceData.replaceFirst(repalceChar, prevOneChar);
                sourceData = replaceDataFromHead(sourceData, repalceChar, index + 1);
            }
            else
            {
                return replaceDataFromTail(sourceData,repalceChar,index);
            }

            break;

        default:{
            prevOneChar = sourceData.substring(index - 1, index);
            prevTwoChar= sourceData.substring(index-2, index-1);
            prevThreeChar= sourceData.substring(index-3, index-2);
            if ((!prevThreeChar.equals(prevTwoChar)|| prevTwoChar.equals(prevOneChar))&&(!prevOneChar.equals(repalceChar)))
            {
                sourceData= sourceData.replaceFirst(repalceChar, prevOneChar);
                sourceData= replaceDataFromHead(sourceData,repalceChar,index+1);
            }
            else
            {
                return replaceDataFromTail(sourceData,repalceChar,index);
            }
        }
    }
    return sourceData;
}
/**
 *  根据规则拆分指定字符串
 * @param sourceData
 * @return
 */
public static String splitData(String sourceData, String repalceChar)
{
    String tmpData= "";
    int startIndex= -1;
    for (int index=0;index< sourceData.length();index++){
        if (sourceData.substring(index,index+1).equals(repalceChar)) {
            sourceData= replaceDataFromHead(sourceData,repalceChar,index);
        }
        if (startIndex==-1) {
            startIndex = index;
            continue;
        }
        if (!sourceData.substring(index,index+1).equals(sourceData.substring(index-1,index)))
        {
            tmpData= tmpData+" "+ sourceData.substring(startIndex,index+1);
            startIndex= -1;
        }
    }
    if (startIndex!= -1)
    {
        tmpData= tmpData+" "+ sourceData.substring(startIndex,sourceData.length());
    }
    return tmpData;
}

Answer 1

没太理解题意，你的代码我执行了一下，没太看明白，
比如输入二进制字符串：111100011010101011，
输出的是 11110 001 10 10 10 10 11，
按我理解应该是：1111 000 11 0 1 0 1 0 1 0 11吧，然后计算其中包含0和1的字符串数量，你先看一下我理解的有问题吗