我正试图从矩阵中提取某个方向的所有对角线，例如右下角。
对于以下矩阵：
A B C D
E F G H
I L M N
预期结果应为
[ [A F M], [B G N], [C H], [D], [E L], [I] ]
欢迎采用一般方法。
我使用的语言是Java。
谢谢！
修改
String[] grid = {"SUGAR",
"GLASS",
"MOUSE"};
for( int k = 0; k < grid.length; k++ )
{
StringBuffer buffer = new StringBuffer( );
for( int i = 0; i < grid.length
&& i+k < grid[0].length( ); i++ )
{
buffer.append( grid[i].charAt(i+k) );
}
trie.addWord( buffer.toString() );
}
添加到trie的输出词是
[ "SLU" "UAS" "GSE" ]
存储在trie中的预期字符串（顺序无关紧要）
[ "SLU" "UAS" "GSE" "GO" "M" "AS" "R"]
答案 0 :(得分：2)
如果您的数据是表格形式，您只需在第一列上扫描矩阵，然后在第一行左侧扫描。
final String[M][N] mtx = { ... };
public List<List<String>> diagonalize() {
final List<List<String>> diags = new ArrayList<>();
for (int row = M - 1; row > 1; --row) {
diags.add(getDiagonal(row, 0));
}
for (int col = 0; col < N; ++col) {
diags.add(getDiagonal(0, col));
}
return diags;
}
private List<String> getDiagonal(int x, int y) {
final List<String> diag = new ArrayList<>();
while (x < M && y < N) {
diag.add(mtx[x++][y++]);
}
return diag;
}
答案 1 :(得分：2)
这是一个有趣的问题需要解决。
很容易陷入嵌套循环中。
我注意到如果把单词放在一个字符串中，就会出现一个模式。
以OP为例，三个词“SUGAR”，“GLASS”，“MOUSE”连接在一起成为SUGARGLASSMOUSE。
以下是我需要从连接字符串中获取的字符的从零开始的字符位置。我把它们排成一行，这样你就可以更容易地看到这种模式。
10 M
5 11 GO
0 6 12 SLU
1 7 13 UAS
2 8 14 GSE
3 9 AS
4 R
看模式了吗？我有3个索引，包括5次迭代。我有3个单词，由5个字母组成。
对角线的数量为letters + words - 1。我们减1，因为字符位置0的第一个字母只使用一次。
以下是我跑过的测试结果。
[ "SUGAR" "GLASS" "MOUSE" "STATE" "PUPIL" "TESTS" ]
[ "T" "PE" "SUS" "MTPT" "GOAIS" "SLUTL" "UASE" "GSE" "AS" "R" ]
[ "SUGAR" "GLASS" "MOUSE" ]
[ "M" "GO" "SLU" "UAS" "GSE" "AS" "R" ]
这是代码：
import java.util.ArrayList;
import java.util.List;
public class Matrix {
public static final int DOWN_RIGHT = 1;
public static final int DOWN_LEFT = 2;
public static final int UP_RIGHT = 4;
public static final int UP_LEFT = 8;
public String[] getMatrixDiagonal(String[] grid, int direction) {
StringBuilder builder = new StringBuilder();
for (String s : grid) {
builder.append(s);
}
String matrixString = builder.toString();
int wordLength = grid[0].length();
int numberOfWords = grid.length;
List<String> list = new ArrayList<String>();
if (wordLength > 0) {
int[] indexes = new int[numberOfWords];
if (direction == DOWN_RIGHT) {
indexes[0] = matrixString.length() - wordLength;
for (int i = 1; i < numberOfWords; i++) {
indexes[i] = indexes[i - 1] - wordLength;
}
int wordCount = numberOfWords + wordLength - 1;
for (int i = 0; i < wordCount; i++) {
builder.delete(0, builder.length());
for (int j = 0; (j <= i) && (j < numberOfWords); j++) {
if (indexes[j] < wordLength * (wordCount - i)) {
char c = matrixString.charAt(indexes[j]);
builder.append(c);
indexes[j]++;
}
}
String s = builder.reverse().toString();
list.add(s);
}
}
if (direction == DOWN_LEFT) {
// Exercise for original poster
}
if (direction == UP_RIGHT) {
// Exercise for original poster
}
if (direction == UP_LEFT) {
// Exercise for original poster
// Same as DOWN_RIGHT with the reverse() removed
}
}
return list.toArray(new String[list.size()]);
}
public static void main(String[] args) {
String[] grid1 = { "SUGAR", "GLASS", "MOUSE", "STATE", "PUPIL", "TESTS" };
String[] grid2 = { "SUGAR", "GLASS", "MOUSE" };
Matrix matrix = new Matrix();
String[] output = matrix.getMatrixDiagonal(grid1, DOWN_RIGHT);
System.out.println(createStringLine(grid1));
System.out.println(createStringLine(output));
output = matrix.getMatrixDiagonal(grid2, DOWN_RIGHT);
System.out.println(createStringLine(grid2));
System.out.println(createStringLine(output));
}
private static String createStringLine(String[] values) {
StringBuilder builder = new StringBuilder();
builder.append("[ ");
for (String s : values) {
builder.append("\"");
builder.append(s);
builder.append("\" ");
}
builder.append("]");
return builder.toString();
}
}
答案 2 :(得分：2)
String[] grid = {"SUGAR",
"GLASS",
"MOUSE"};
System.out.println("Result: " + Arrays.toString(diagonals(grid)));
public static String[] diagonals(String[] grid) {
int nrows = grid.length;
int ncols = grid[0].length();
int nwords = ncols + nrows - 1;
String[] words = new String[nwords];
int iword = 0;
for (int col = 0; col < ncols; ++col) {
int n = Math.min(nrows, ncols - col);
char[] word = new char[n];
for (int i = 0; i < n; ++i) {
word[i] = grid[i].charAt(col + i);
}
words[iword] = new String(word);
++iword;
}
for (int row = 1; row < nrows; ++row) {
int n = Math.min(ncols, nrows - row);
char[] word = new char[n];
for (int i = 0; i < n; ++i) {
word[i] = grid[row + i].charAt(i);
}
words[iword] = new String(word);
++iword;
}
assert iword == nwords;
return words;
}
Result: [SLU, UAS, GSE, AS, R, GO, M]
首先是一个包含列上第一个元素的循环。然后在行上循环，跳过第0行。 两个循环中的代码都非常对称。 没什么太难的。假设所有字符串都具有相同的长度。
作为一个循环：
public static String[] diagonals(String[] grid) {
int nrows = grid.length;
int ncols = grid[0].length();
int nwords = ncols + nrows - 1;
String[] words = new String[nwords];
// Position of first letter in word:
int row = 0;
int col = ncols - 1;
for (int iword = 0; iword < nwords; ++iword) {
int n = Math.min(nrows - row, ncols - col);
char[] word = new char[n];
for (int i = 0; i < n; ++i) {
word[i] = grid[row + i].charAt(col + i);
}
words[iword] = new String(word);
if (col > 0) {
--col;
} else {
++row;
}
}
return words;
}
word的声明可能会被置于循环之外。只需走（左，上）左边和上边缘。
答案 3 :(得分：0)
您可以使用二维数组
表示矩阵char [] [] matrix = char [] []
然后你可以使用for循环来迭代它并提取你想要的输出，你的算法的输入将是你想要的对角线方向。
对于Ex;一个可能的输入将是正确的
根据可能的输入，您必须决定如何迭代循环初始条件和终止条件。
从最后一列第一行字符
开始r = 0，c = coulmn_count -1;
终止条件将是first_column最后一行索引处的字符。
r = row_count -1，c = 0;
在最后一行或最后一列读取字符时的每次迭代都是子循环的终止