根据词典顺序使用自定义词典（abcdefghijklmnopqrstuvwxyz的排列）对字符串数组进行排序。这是代码。
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import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
/**
*
* @author sabertooth
*/
public class SortString {
/**
* @param args the command line arguments
*/
private static char[] index;
private static BufferedReader br;
public static void main(String[] args) throws Exception {
// TODO code application logic here
br=new BufferedReader(new InputStreamReader(System.in));
int testCases=Integer.parseInt(br.readLine());
for(int i=0;i<testCases;i++)
{
String dictionary=br.readLine();
index=new char[dictionary.length()];
index=dictionary.toCharArray();
int set=Integer.parseInt(br.readLine());
String[] unsortedInput=new String[set];
String[] sortedInput=new String[set];
for(int j=0;j<set;j++)
{
unsortedInput[j]=br.readLine();
}
if(unsortedInput.length<=1)
{
System.out.println(unsortedInput[0]);
}
else
{
// merge sort on this array
sortedInput=mergeSort(unsortedInput);
for(int k=0;k<sortedInput.length;k++)
{
System.out.println(sortedInput[k]);
}
}
}
}
private static String[] mergeSort(String[] unsortedInput){
if(unsortedInput.length<=1)
{
return unsortedInput;
}
String[] left;
String[] right;
int middle=unsortedInput.length/2;
if(unsortedInput.length%2==0)
{
left=new String[middle];
right=new String[middle];
}
else
{
left=new String[middle];
right=new String[middle+1];
}
System.arraycopy(unsortedInput, 0, left, 0, middle);
System.arraycopy(unsortedInput, middle, right, 0, unsortedInput.length - middle);
left=mergeSort(left);
right=mergeSort(right);
return merge(left,right);
}
private static String[] merge(String[] left, String[] right){
List<String> leftList=new ArrayList<String>();
List<String> rightList=new ArrayList<String>();
List<String> result=new ArrayList<String>();
leftList.addAll(Arrays.asList(left));
rightList.addAll(Arrays.asList(right));
while(leftList.size()>0 || rightList.size()>0)
{
if(leftList.size()>0 && rightList.size()>0)
{
// my own comparison
if(compare(leftList.get(0),rightList.get(0))==-1)
{
// leftString is less than right string
result.add(leftList.get(0));
leftList=leftList.subList(1, leftList.size());
}
else if(compare(leftList.get(0),rightList.get(0))==1)
{
//left string is greater than right string
result.add(rightList.get(0));
rightList=rightList.subList(1, rightList.size());
}
else if(compare(leftList.get(0),rightList.get(0))==0)
{
// leftString is equal to right string
result.add(leftList.get(0));
leftList=leftList.subList(1, leftList.size());
}
}
else if(leftList.size()>0)
{
for(int i=0;i<leftList.size();i++)
{
result.add(leftList.get(i));
}
leftList.clear();
}
else if(rightList.size()>0)
{
for(int i=0;i<rightList.size();i++)
{
result.add(rightList.get(i));
}
rightList.clear();
}
}
String[] sortedInput=new String[result.size()];
for(int i=0;i<result.size();i++)
{
sortedInput[i]=result.get(i);
}
return sortedInput;
}
private static int compare(String leftString, String rightString){
// return -1 if left string is less than right string else left string is greater than right string return 1
int min=Math.min(leftString.length(), rightString.length());
int response=0;
for(int i=0;i<min;i++)
{
if(compareChar(leftString.charAt(i), rightString.charAt(i))==-1)
{
response= -1;
break;
}
else if(compareChar(leftString.charAt(i), rightString.charAt(i))==1)
{
response= 1;
break;
}
else if(compareChar(leftString.charAt(i), rightString.charAt(i))==0)
{
response=0;
}
}
return response;
}
private static int compareChar(char x,char y){
// returns true if x <y
int indexofx=0;
int indexofy=0;
int response=0;
for(int i=0;i<index.length;i++)
{
if(index[i]==x)
{
indexofx=i;
}
if(index[i]==y)
{
indexofy=i;
}
}
if(indexofx<indexofy)
{
response=-1;
}
else if(indexofx>indexofy)
{
response=1;
}
else if(indexofx==indexofy)
{
response=0;
}
return response;
}
}
问题是当我为某些输入运行此输出时输出正确，而对于其他输出则输出不正确。我一直在调试它，但无法找到错误。
编辑：
Adriana was playing with the English alphabet. When she was done playing with the alphabet, she realised that she had jumbled up the positions of the letters. Now, given a set of words, she wondered what would be the dictionary ordering of these words based on the new alphabet ordering which she made.
In other words, given a permutation of the English alphabet, E and a set of words S, you need to output the lexicographical ordering of the words in the set S based on the new alphabet, E.
Input
The first line will contain a single integer, T, denoting the number of test cases. T lines follow.
For each test case:
The first line will contain a string, E, the new alphabet ordering, which will be a permutation of 'abcdefghijklmnopqrstuvwxyz'
The next line will contain a single integer M, the size of the set S. S lines follow, each containing a single word, containing lowercase latin characters.
Output
For each test case, output S lines, each line containing one word from the set S, ordered lexicographically.
Constraints
1 <= T <= 1000
1 <= M <= 100
1 <= |W| <= 50
Sample Input
2
abcdefghijklmnopqrstuvwxyz
2
aa
bb
bacdefghijklmnopqrstuvwxyz
2
aa
bb
Sample Output
aa
bb
bb
aa
答案 0 :(得分：0)
这适用于任何给定的排列。为什么使用自定义排序函数（除非你必须对自定义类对象进行排序）当java为你提供内置函数时？
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.ArrayList;
class TestClass {
public static ArrayList<String> res;
static void sort(String dictionary, String[] words, int count){
String eng = "abcdefghijklmnopqrstuvwxyz";
String[] tempArray = new String[count];
String temp="";
char ch;
int index, m=0;
for(String x : words){
temp = "";
for(int l =0 ;l<x.length(); l++){
ch = x.charAt(l);
index = dictionary.indexOf(ch);
temp = temp + eng.charAt(index);
}
tempArray[m] = temp;
m++;
}
Arrays.sort(tempArray);
for(String x : tempArray){
temp = "";
for(int l =0 ;l<x.length(); l++){
ch = x.charAt(l);
index = eng.indexOf(ch);
temp = temp + dictionary.charAt(index);
}
res.add(temp);
}
}
public static void main(String args[] ) throws Exception {
res = new ArrayList<String>();
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
int N = Integer.parseInt(line);
int count;
String [] words;
String dictionary;
for (int i = 0; i < N; i++) {
dictionary = br.readLine();
count = Integer.parseInt(br.readLine());
words = new String[count];
for(int j =0; j<count; j++){
words[j] = br.readLine();
}
sort(dictionary,words,count);
}
for(String x : res)
System.out.println(x);
}
}
我想这是某种算法挑战。是吗？ ：）