One measure of ``unsortedness'' in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequence ``DAABEC'', this measure is 5, since D is greater than four letters to its right and E
is greater than one letter to its right. This measure is called the number of inversions in the sequence. The sequence ``AACEDGG'' has only one inversion (E and D)---it is nearly sorted---while the sequence ``ZWQM'' has 6 inversions (it is as unsorted as can
be---exactly the reverse of sorted).
You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of ``sortedness'', from ``most sorted'' to ``least sorted''.
All the strings are of the same length.
Input
The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (0 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.
Output
Output the list of input strings, arranged from ``most sorted'' to ``least sorted''. Since two strings can be equally sorted, then output them according to the orginal order.
二、问题分析
题意：求输入的一串字母的秩（字母序的规整程度），并按秩由小到大输出字符串。此题的难点不在于求秩，而是再求的秩之后怎么把它和字符串管连起来。由于求得秩之后还要把秩按升序排序，所以关联字符串就比较麻烦了。
三、问题解决
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一开始看到要关联两个值就想到了Map,于是就把秩的值存为键，把字符串存为值。发现要排序，于是就用了TreeMap来自动维护键的顺序。一切进展顺利，以为又轻松解掉一题。刚开始运行给定测试数据的时候，没有问题过了。但是一提交代码，悲剧啊，WA.为什么呢？于是又找了几组测试数据，发现有的测试数据的输出比输入的行数要少。几经测试，发现了一个很容易被忽视的问题，那就是几个字符串的秩很可能是相同的，而我们亲爱的Map的键必须是独一无二的，重复添加相同的键的键值对，Map是不会接受的。好的我刚开始的尝试就这样失败了。
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后来一想，既然键必须唯一，那我就把原来的键值对交换一下了，把秩存入值，用字符串来做键。但又有一个问题，就是如果存在相同的秩，在查找键的过程中，会找到多个键。好像还是不行，但是还是可行的。在遇到有相同的键的情况下，每找到对应值的键后，就把键值对移除，那么在下一个相同秩进行查找是就不会出现上次查找的键了。这样做的过程比较麻烦，还要把秩从Map中分离出来，然后进行排序。之后又要把秩和Map中的值进行关联判断，再输出结果。虽然程序可以运行，并输出结果，但是提交上去发现Runtime error.
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参考了大神的方法，用Java的面向对象的方法可以巧妙解决这个关联的问题。把每个字符串看做是一个对象，把秩看出是它的一个属性。而计算秩就放到类中进行。这样就把字符串和秩关联起来了。再用容器把每个DNA对象添加。之后调用Collections.sort(Collection,new compartor);来排序。这里用到了实现了Compartor接口的类，它是强行对某个对象 collection 进行整体排序 的比较函数。可以将 Comparator 传递给 sort 方法（如 或 ），从而允许在排序顺序上实现精确控制。排序之后，只要打印容器即可。应用语言特性来解决这个问题，小弟对您的敬仰如滔滔江水，绵绵不绝；又如黄河之水，一发不可收拾。
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Scanner;
class DNA{
private int sortNum = 0;
public String str = null;
public DNA(String str){
this.str=str;
int num = 0;
char[] c=str.toCharArray();
for(int i=0 ;i < c.length - 1 ;i++){
for(int j=i+1 ;j < c.length ;j++){
if(c[i] > c[j])
num++;
}
}
sortNum=num;
}
public int getSortNum(){
return sortNum;
}
public String toString(){
return str;
}
}
class DNACompartor implements Comparator{
public int compare(Object o1, Object o2) {
DNA a=(DNA) o1;
DNA b=(DNA) o2;
if(a.getSortNum() > b.getSortNum())
return 1;
else if(a.getSortNum() == b.getSortNum())
return 0;
else
return -1;
}
}
public class N1007_DNASorting_Version2{
public static void main(String args[]){
Scanner cin =new Scanner(System.in);
int n =cin.nextInt();
int t =cin.nextInt();
ArrayList<DNA> list=new ArrayList<DNA>();
for(int i=0 ;i < t; i++){
DNA a=new DNA(cin.next());
list.add(a);
}
Collections.sort(list, new DNACompartor());
for(DNA s:list){
System.out.println(s.toString());
}
}
}
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