我正在寻找二维数组中的单词，我得到了搜索作品（垂直，水平和对角线）
我只是无法以这种特定格式打印找到的单词。就像它打印实际上在2D数组中找到它的单词。 我希望我的想法很明确
代码从file.txt读取Charters并将它们保存到2d数组中。 我们在这个数组中搜索一个单词。
并尝试按照他们所在的顺序找到它们。 我附上了一个例子
Puzzle word file Output
n o h t y p s ruby
m i a r y c c cave c
l l e k s a h a
r u b y v m e r u b y v
e h h e l l m e
p c j n i c e
r e e k b i p
这是代码......
import java.awt.Point;
import java.io.File;
import java.io.FileNotFoundException;
import java.util.HashSet;
import java.util.LinkedHashSet;
import java.util.Scanner;
import java.util.Set;
public class WordPuzzle {
// to print..
private Set<Point> wordPoints = new HashSet<>();
public Set<String> findWords(char[][] puzzle, String[] inputWords) {
Set<String> foundWords = new HashSet<String>();
int minimumWordLength = findMinimumWordLength(inputWords);
Set<String> possibleWords = findPossibleWords(puzzle, minimumWordLength);
for(String word : inputWords) {
for(String possibleWord : possibleWords) {
if(possibleWord.contains(word) || possibleWord.contains(new StringBuffer(word).reverse())) {
foundWords.add(word);
break;
}
}
}
return foundWords;
}
private int findMinimumWordLength(String[] inputWords) {
int minimumLength = Integer.MAX_VALUE;
for(String word : inputWords) {
if(word.length() < minimumLength)
minimumLength = word.length();
}
return minimumLength;
}
private Set<String> findPossibleWords(char[][] puzzle, int minimumWordLength) {
Set<String> possibleWords = new LinkedHashSet<String>();
int dimension = puzzle.length; //Assuming puzzle is square
if(dimension >= minimumWordLength) {
/* Every row in the puzzle is added as a possible word holder */
for(int i = 0; i < dimension; i++) {
if(puzzle[i].length >= minimumWordLength) {
possibleWords.add(new String(puzzle[i]));
}
}
/* Every column in the puzzle is added as a possible word holder */
for(int i = 0; i < dimension; i++) {
StringBuffer temp = new StringBuffer();
for(int j = 0; j < dimension; j++) {
temp = temp.append(puzzle[j][i]);
}
possibleWords.add(new String(temp));
}
/* Adding principle diagonal word holders */
StringBuffer temp1 = new StringBuffer();
StringBuffer temp2 = new StringBuffer();
for(int i = 0; i < dimension; i++) {
temp1 = temp1.append(puzzle[i][i]);
temp2 = temp2.append(puzzle[i][dimension - i - 1]);
}
possibleWords.add(new String(temp1));
possibleWords.add(new String(temp2));
/* Adding non-principle diagonal word holders */
for(int i = 1; i < dimension - minimumWordLength; i++) {
temp1 = new StringBuffer();
temp2 = new StringBuffer();
StringBuffer temp3 = new StringBuffer();
StringBuffer temp4 = new StringBuffer();
for(int j = i, k = 0; j < dimension && k < dimension; j++, k++) {
temp1 = temp1.append(puzzle[j][k]);
temp2 = temp2.append(puzzle[k][j]);
temp3 = temp3.append(puzzle[dimension - j - 1][k]);
temp4 = temp4.append(puzzle[dimension - k - 1][j]);
}
possibleWords.add(new String(temp1));
possibleWords.add(new String(temp2));
possibleWords.add(new String(temp3));
possibleWords.add(new String(temp4));
}
}
return possibleWords;
}
public static void main(String args[]) throws FileNotFoundException {
WordPuzzle program = new WordPuzzle();
////////////////////////////////////////// if you want to copy the word file, copy here. //////
// allocate the file..
File inFile = new File("C:\\Users\\admin\\Desktop\\workspace\\DS600\\src\\Puzzle.txt");
// scan the file..
Scanner scanner = new Scanner(inFile);
// create an array of type char
char[][] puzzleInput = new char[50][50];
//assign every letter in the char array
for(int i=0; i < 50; i++) {
puzzleInput[i] = scanner.nextLine().toCharArray();
}
scanner.close();
////////////////////////// end the copy file here. //////////////////////////
String[] InputWords = {"ONE", "TWO","THREE","POLAND"};
Set<String> wordsFound = program.findWords(puzzleInput, InputWords);
// PRINT OUT... ONLY
for (int y = 0; y < puzzleInput.length; y++) {
for (int x = 0; x < puzzleInput[y].length; x++) {
// HOW TO PRINT IT .. ??
}
System.out.println();
}
/*
for(String word : wordsFound) {
System.out.println(word);
}
*/
}
}
答案 0 :(得分：0)
你可以做的是有一个第二个数组，它只是空的，但与第一个数组的尺寸相同，每次在第一个数组中找到一个单词时，你都会按顺序将单词中的所有字母添加到第二个数组中第一个。因此，例如，一旦您查看第一个数组并在某个位置找到“ruby”，请将“ruby”添加到第二个数组的相同位置。这样，您将所有找到的单词添加到空数组中，从而生成一个只包含您在完全相同位置找到的单词的数组。
答案 1 :(得分：0)
这是我在编码竞赛中解决类似问题时编写的工作代码示例。它在2-D char数组中找到单词并用坐标打印出来。希望这会有所帮助......
public class Solution {
public static void main(String[] args) {
char[][] crossword = new char[][]{
{'f', 'd', 'e', 'r', 'l', 'k'},
{'u', 's', 'a', 'm', 'e', 'o'},
{'l', 'n', 'g', 'r', 'o', 'v'},
{'m', 'l', 'p', 'r', 'r', 'h'},
{'p', 'o', 'e', 'e', 'j', 'j'}
};
List<Word> allWords = detectAllWords(crossword, "home", "same");
for(Word word : allWords) {
System.out.println(word.toString());
}
/*
Expected result:
home - (5, 3) - (2, 0)
same - (1, 1) - (4, 1)
*/
}
public static List<Word> detectAllWords(char[][] crossword, String... words) {
List<Word> wordList = new ArrayList<>();
int[][] searchDirections = new int[][] {
{0, 1},
{1, 1},
{1, 0},
{1, -1},
{0, -1},
{-1, -1},
{-1, 0},
{-1, 1},
};
for (String word : words) nextWord:{
for (int i = 0; i < crossword.length; i++) {
for (int j = 0; j < crossword[i].length; j++) {
if (word.charAt(0) == crossword[i][j]) {
for (int directions = 0; directions < searchDirections.length; directions++) {
int tmp_i = i, tmp_j = j, wordPos = 1;
while (wordPos < word.length()) {
tmp_i += searchDirections[directions][0];
tmp_j += searchDirections[directions][1];
if (tmp_i < 0 || tmp_i >= crossword.length || tmp_j < 0 || tmp_j >= crossword[tmp_i].length)
break;
if (word.charAt(wordPos) != crossword[tmp_i][tmp_j])
break;
else if (wordPos == word.length() - 1) {
Word tWord = new Word(word);
tWord.setStartPoint(j, i);
tWord.setEndPoint(tmp_j, tmp_i);
wordList.add(tWord);
break nextWord;
}
wordPos++;
}
}
}
}
}
}
return wordList;
}
public static class Word {
private String text;
private int startX;
private int startY;
private int endX;
private int endY;
public Word(String text) {
this.text = text;
}
public void setStartPoint(int i, int j) {
startX = i;
startY = j;
}
public void setEndPoint(int i, int j) {
endX = i;
endY = j;
}
@Override
public String toString() {
return String.format("%s - (%d, %d) - (%d, %d)", text, startX, startY, endX, endY);
}
}
}
答案 2 :(得分：0)
除了我原来的答案，这里是您完全需要的完整代码。请注意print2DArray（...）静态方法，该方法执行将找到的单词打印在与原始数组内相同位置的所有工作：
public class Solution {
public static void main(String[] args) {
char[][] crossword = new char[][]{
{'f', 'd', 'e', 'r', 'l', 'k'},
{'u', 's', 'a', 'm', 'e', 'o'},
{'l', 'n', 'g', 'r', 'o', 'v'},
{'m', 'l', 'p', 'r', 'r', 'h'},
{'p', 'o', 'e', 'e', 'j', 'j'}
};
List<Word> allWords = detectAllWords(crossword, "home", "same");
// Print the results with starting and ending coordinates:
for(Word word : allWords) {
System.out.println(word.toString());
}
// Print the results as they are inside the original array
print2DArray(allWords, crossword);
}
public static void print2DArray(List<Word> words, char[][] array) {
char[][] toPrint = new char[array.length][array[0].length];
for (int i = 0; i < array.length; i++) {
for (int j = 0; j < array[i].length; j++) {
toPrint[i][j] = '#';
}
}
for(Word word : words) {
int vectorX = word.endX - word.startX;
int vectorY = word.endY - word.startY;
if(vectorX > 0) {
vectorX = 1;
} else if(vectorX < 0) {
vectorX = -1;
}
if(vectorY > 0) {
vectorY = 1;
} else if(vectorY < 0) {
vectorY = -1;
}
for(int i = 0; i < word.text.length(); i++) {
toPrint[word.startY + (i * vectorY)][word.startX + (i * vectorX)] = word.text.charAt(i);
}
}
for (int i = 0; i < toPrint.length; i++) {
for (int j = 0; j < toPrint[i].length; j++) {
if(toPrint[i][j] == '#') {
if(j == toPrint[i].length -1) {
System.out.println(" ");
} else {
System.out.print(" ");
}
} else {
if(j == toPrint[i].length -1) {
System.out.println(" " + toPrint[i][j]);
} else {
System.out.print(" " + toPrint[i][j]);
}
}
}
}
}
public static List<Word> detectAllWords(char[][] crossword, String... words) {
List<Word> wordList = new ArrayList<>();
int[][] searchDirections = new int[][]{
{0, 1},
{1, 1},
{1, 0},
{1, -1},
{0, -1},
{-1, -1},
{-1, 0},
{-1, 1},
};
for (String word : words)
nextWord:{
for (int i = 0; i < crossword.length; i++) {
for (int j = 0; j < crossword[i].length; j++) {
if (word.charAt(0) == crossword[i][j]) {
for (int directions = 0; directions < searchDirections.length; directions++) {
int tmp_i = i, tmp_j = j, wordPos = 1;
while (wordPos < word.length()) {
tmp_i += searchDirections[directions][0];
tmp_j += searchDirections[directions][1];
if (tmp_i < 0 || tmp_i >= crossword.length || tmp_j < 0 || tmp_j >= crossword[tmp_i].length)
break;
if (word.charAt(wordPos) != crossword[tmp_i][tmp_j])
break;
else if (wordPos == word.length() - 1) {
Word tWord = new Word(word);
tWord.setStartPoint(j, i);
tWord.setEndPoint(tmp_j, tmp_i);
wordList.add(tWord);
break nextWord;
}
wordPos++;
}
}
}
}
}
}
return wordList;
}
public static class Word {
private String text;
private int startX;
private int startY;
private int endX;
private int endY;
public Word(String text) {
this.text = text;
}
public void setStartPoint(int i, int j) {
startX = i;
startY = j;
}
public void setEndPoint(int i, int j) {
endX = i;
endY = j;
}
@Override
public String toString() {
return String.format("%s - (%d, %d) - (%d, %d)", text, startX, startY, endX, endY);
}
}
}