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Solution1.java
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Solution1.java
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package test.didi;
import org.junit.Test;
import java.sql.Time;
import java.util.Random;
/**@https://labuladong.gitbook.io/algo/gao-pin-mian-shi-xi-lie/k-ge-yi-zu-fan-zhuan-lian-biao
* 要的是对链表的改函数的!
* */
public class Solution1 {
//可以指定字符串的某个位置是什么范围的值
@Test
public void testString() {
String s = getRandomString2(29);
System.out.println(s);
}
public String getRandomString2(int length){
Random random=new Random();
StringBuffer sb=new StringBuffer();
for(int i=0;i<length;i++){
int number= 0;
try {
number = random.nextInt(3);//0-2的生成。
} catch (Exception e) {
e.printStackTrace();
}
long result=0;
switch(number){
case 0:
result=Math.round(Math.random()*25+65);
sb.append(String.valueOf((char)result));
break;
case 1:
result=Math.round(Math.random()*25+97);
sb.append(String.valueOf((char)result));
break;
case 2:
sb.append(String.valueOf(new Random().nextInt(10)));
break;
}
}
return sb.toString();
}
public String reverse(int start, int end, String s) {
StringBuilder stringBuilder = new StringBuilder();
while(start != end) {
/*在这里面进行string 的反转*/
stringBuilder.insert(0, s.charAt(start));
start++;
}
return stringBuilder.toString();
}
public String reverseK(String s, int group) {
if (s == null) {
return null;
}
StringBuilder stringBuilder = new StringBuilder();
StringBuilder stringBuilder1 = new StringBuilder();
int groups = s.length()-s.length()%group;
int index = 0, pos = 0;
while(index < groups) {
for (int i = 0; i < group; i++) {
/**找到b的位置, 我不要这个for循环时可以的吗?
* @每次需要加多少呢?对吧.需要的!!!
*
* */
index++;
}
String temp = reverse(pos, index, s);//多个的temp怎么连接?
stringBuilder.append(temp);
pos = index;
}
for (int i = groups; i < s.length(); i++) {
stringBuilder1.insert(0, s.charAt(i));
}
stringBuilder.append(stringBuilder1);
return stringBuilder.toString();
}
@Test
public void test(){
long time = System.currentTimeMillis();
String s = getRandomString2(79);
System.out.println(s);
System.out.println(reverseK(s,5));
/*100个char。 2 ^ 8*100*1000 ~= 10^8的处理速度
*
* */
long end = System.currentTimeMillis();
System.out.println((float) (end- time)/1000 + "s");
}
}