获取两个字符串中的相同子串的一些残留问题 - 黑马程序员技术交流社区

class StringTest
{
//主函数
public static void main(String[] args)
{
//String str = "abghigfdkljtsawddcdefdddddddddddeddd";
//String str1 = "d cdef c kljt d tsaw b fdkl";
String str = "abcdefghigklmnkhjtfdsaw";
String str1 = "d bcdefghi c cdefghig";
getMaxString(str,str1);
}
//打印功能
public static void sop(Object str)
{
System.out.println(str);
}
public static void getMaxString(String str1,String str2)
{
思路：比较str1与str2长度，用短的那个字符串来一次递减长度，看长的那个是否包含短的那个字符串
boolean contains()
for循环中，三个变量，numOfChar，beginIndex，endIndex 的含义
x代表字符串2减少的字符数
beginIndex 代表减少字符后获取子串的起始坐标
endIndex 代表减少字符后获取子串的结束坐标
String min ="";
String max = "";
int flag = 0;//用于获取首次遇到相同子串的时候，减少的字符数。之后把它用于循环中检查减少该字符数后是否还有其他相同子串。
if(str1.length()>str2.length())
{
min = str2;
max = str1;
}
out:for(int numOfChar = 0;numOfChar<min.length();numOfChar++)
{
flag = numOfChar;
//sop(flag);
for(int beginIndex = 0,endIndex=min.length()-numOfChar;endIndex<min.length()+1;)
{
//if(max.contains(min.substring(beginIndex,endIndex)))
//sop(min.substring(beginIndex,endIndex));
if(max.contains(min.substring(beginIndex,endIndex)))
{
min.substring(beginIndex,endIndex);
break out;
}
beginIndex++;
endIndex++;
}
}
for(int beginIndex = 0,endIndex=min.length()-flag;endIndex<min.length()+1;)
{
if(max.contains(min.substring(beginIndex,endIndex)))
{
sop(min.substring(beginIndex,endIndex));
}
beginIndex++;
endIndex++;
}
}
}

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public ArrayList<String> getLonggestPubSubStr(String shortter,String longger){
ArrayList<String> tempList=new ArrayList<String>();
int length=shortter.length();
for(int len=length;len>0;len--){
for(int i=0;i<length-len+1;i++){
String tempStr=shortter.substring(i, len+i);
if(longger.indexOf(tempStr)!=-1){
tempList.add(tempStr);
}
}
if(tempList.size()!=0)
return tempList;
}
return null;
}

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String str = "abcdefghigklmnkhjtfdsaw";
String str1 = "d bcdefghi c cdefghig";
System.out.println(getLonggestPubSubStr(str, str1));

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class TestDemo {
public static void main(String argv[]) {
String str = "abcdefghigklmnkhjtfdsaw";
String subStr = "d bcdefghi c cdefghig";
System.out.println(getCommonStr(str, subStr));
}
public static List getCommonStr(String str, String subStr) {
List<String> list = new ArrayList<String>();
for(int i=0; i< subStr.length(); i++) {
for(int j=0,m=subStr.length()-i; m!=subStr.length()+1; j++,m++) {
if(str.contains(subStr.substring(j, m)))
list.add(subStr.substring(j, m));
}
if(list.size() != 0)
return list;
}
return null;
}
}

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