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2013
11-08

POJ 1007 DNA Sorting [解题报告] Java

DNA Sorting

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

样例输出:

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

解题代码:

import java.util.*;   
  
class DNA   
{   
    private String str = null;   
    private int sortNum = 0;   
       
    public DNA(String input)   
    {   
        str = input;   
           
        int num = 0;   
        for(int i = 0; i < str.length()-1; i++)   
        {   
            for(int j = i+1; j < str.length(); j++)   
                if(str.charAt(i) > str.charAt(j))   
                    num++;   
        }   
        sortNum = num;   
    }   
       
    public int getSortNum()   
    {   
        return sortNum;   
    }   
       
    public String toString()   
    {   
        return str;   
    }   
}   
  
class DNAComparator implements Comparator   
{   
    public int compare(Object o1, Object o2)   
    {   
        DNA d1 = (DNA)o1;   
        DNA d2 = (DNA)o2;   
           
        if(d1.getSortNum() > d2.getSortNum())   
            return 1;   
        else if(d1.getSortNum() == d2.getSortNum())   
            return 0;   
        else  
            return -1;   
    }   
}   
  
public class Main {   
  
    public static void main(String[] args) {   
        Scanner cin = new Scanner(System.in);   
        String[] str = cin.nextLine().split(" ");   
           
        int col = Integer.valueOf(str[0]).intValue();   
        int row = Integer.valueOf(str[1]).intValue();   
        List list = new ArrayList();   
           
        for(int i = 0; i < row; i++)   
        {   
            DNA dna = new DNA(cin.nextLine());   
            list.add(dna);   
        }   
           
        Collections.sort(list, new DNAComparator());   
        print(list);   
    }   
       
    private static void print(List list)   
    {   
        Iterator iter = list.iterator();   
        while(iter.hasNext())   
        {   
            System.out.println(iter.next());   
        }   
    }   
  
}

  1. 其实国内大部分公司对算法都不够重视。特别是中小型公司老板根本都不懂技术,也不懂什么是算法,从而也不要求程序员懂什么算法,做程序从来不考虑性能问题,只要页面能显示出来就是好程序,这是国内的现状,很无奈。

  2. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。