leetcode :937. 重新排列日志文件

迈不过友情╰ 2021-12-12 15:55 176阅读 0赞

### leetcode :937. 重新排列日志文件 ###

你有一个日志数组 logs。每条日志都是以空格分隔的字串。

对于每条日志，其第一个字为字母数字标识符。然后，要么：

标识符后面的每个字将仅由小写字母组成，或；  
标识符后面的每个字将仅由数字组成。  
我们将这两种日志分别称为字母日志和数字日志。保证每个日志在其标识符后面至少有一个字。

将日志重新排序，使得所有字母日志都排在数字日志之前。字母日志按内容字母顺序排序，忽略标识符；在内容相同时，按标识符排序。数字日志应该按原来的顺序排列。

返回日志的最终顺序。

示例 ：

> 输入：\[“a1 9 2 3 1”,“g1 act car”,“zo4 4 7”,“ab1 off key dog”,“a8 act zoo”\]  
> 输出：\[“g1 act car”,“a8 act zoo”,“ab1 off key dog”,“a1 9 2 3 1”,“zo4 4 7”\]

提示：

1. 0 <= logs.length <= 100
2. 3 <= logs\[i\].length <= 100
3. logs\[i\] 保证有一个标识符，并且标识符后面有一个字。

解决方案：

package string;
 
 import com.sun.org.apache.bcel.internal.generic.NEW;
 
 import java.util.ArrayList;
 import java.util.Collections;
 import java.util.List;
 
 public class ReorderLogFiles_zane_0703 {
 
 
 public static void main(String[] args) {
 ReorderLogFiles_zane_0703 reorderLogFiles_zane_0703 = new ReorderLogFiles_zane_0703();
 String[] record = {
 "a1 9 2 3 1","g1 act car","zo4 4 7","ab1 off key dog","a8 act zoo"
 };
 String[] array = reorderLogFiles_zane_0703.reorderLogFiles(record);
 for (String a : array) {
 System.out.println(a);
 }
 }
 /**
 * No.937
 * 解题思路：
 * 新建一个Log内部类并实现compareable接口，有标识符和日志内容两个属性，重写compareTo方法
 * 根据每个字符串的最后一个字符来判断该字符串是否是字母日志还是数字日志
 * 将字母日志和数字日志分开存放。
 * 对字母日志进行排序。
 * 将排序后的字母日志和数字日志进行合并。
 * 将list转为字符串数组返回。
 * @param logs
 * @return
 */
 public String[] reorderLogFiles(String[] logs) {
 
 List<Log> alist = new ArrayList<>();
 List<String> numList = new ArrayList<>();
 
 
 for (String log : logs) {
 int length = log.length();
 
 if (log.charAt(length - 1) >= 'a' && log.charAt(length - 1) <= 'z') {
 int index = log.indexOf(' ');
 Log alog = new Log();
 alog.setFlag(log.substring(0,index));
 alog.setTail(log.substring(index, log.length()));
 alist.add(alog);
 }else {
 numList.add(log);
 }
 }
 Collections.sort(alist);
 for (int i = alist.size() - 1; i >= 0; i--) {
 Log log = alist.get(i);
 String astring = log.getFlag() + log.getTail();
 numList.add(0, astring);
 }
 
 return (String[]) numList.toArray(new String[numList.size()]);
 }
 
 
 class Log implements Comparable {
 private String flag;
 private String tail;
 
 public Log() {
 }
 
 public Log(String flag, String tail) {
 this.flag = flag;
 this.tail = tail;
 }
 
 
 public String getFlag() {
 return flag;
 }
 
 public void setFlag(String flag) {
 this.flag = flag;
 }
 
 public String getTail() {
 return tail;
 }
 
 public void setTail(String tail) {
 this.tail = tail;
 }
 
 /**
 * Compares this object with the specified object for order. Returns a
 * negative integer, zero, or a positive integer as this object is less
 * than, equal to, or greater than the specified object.
 *
 * The implementor must ensure <tt>sgn(x.compareTo(y)) ==
 * -sgn(y.compareTo(x))</tt> for all <tt>x</tt> and <tt>y</tt>. (This
 * implies that <tt>x.compareTo(y)</tt> must throw an exception iff
 * <tt>y.compareTo(x)</tt> throws an exception.)
 *
 * The implementor must also ensure that the relation is transitive:
 * <tt>(x.compareTo(y)>0 && y.compareTo(z)>0)</tt> implies
 * <tt>x.compareTo(z)>0</tt>.
 *
 * Finally, the implementor must ensure that <tt>x.compareTo(y)==0</tt>
 * implies that <tt>sgn(x.compareTo(z)) == sgn(y.compareTo(z))</tt>, for
 * all <tt>z</tt>.
 *
 * It is strongly recommended, but not strictly required that
 * <tt>(x.compareTo(y)==0) == (x.equals(y))</tt>. Generally speaking, any
 * class that implements the <tt>Comparable</tt> interface and violates
 * this condition should clearly indicate this fact. The recommended
 * language is "Note: this class has a natural ordering that is
 * inconsistent with equals."
 *
 * In the foregoing description, the notation
 * <tt>sgn(</tt>expression<tt>)</tt> designates the mathematical
 * signum function, which is defined to return one of <tt>-1</tt>,
 * <tt>0</tt>, or <tt>1</tt> according to whether the value of
 * expression is negative, zero or positive.
 *
 * @param o the object to be compared.
 * @return a negative integer, zero, or a positive integer as this object
 * is less than, equal to, or greater than the specified object.
 * @throws NullPointerException if the specified object is null
 * @throws ClassCastException if the specified object's type prevents it
 * from being compared to this object.
 */
 @Override
 public int compareTo(Object o) {
 Log temp = (Log) o;
 if (this.tail.equals(temp.getTail())) {
 return this.flag.compareTo(temp.getFlag());
 }else {
 
 return this.tail.compareTo(temp.getTail());
 }
 }
 
 
 @Override
 public String toString() {
 return "Log{" +
 "flag='" + flag + '\'' +
 ", tail='" + tail + '\'' +
 '}';
 }
 }
 
 }