英文:
Browse the matrix / Lee Algorithm
问题
我必须编写一个方法来遍历这个矩阵。我从键盘输入位置[L，C]的坐标，从这个位置开始进行扩展。只有在下一个值小于当前值时，才会移动到下一个值。不会沿对角线移动！
PS：对于我的英语表示抱歉。
就像图片中所示：
enter image description here
英文:
I have to do a method that will go through the matrix.I give the coordinates from the keyboard of the position [L, C], from where the extension will start.It will pass to the next value only if the next value is smaller than this.On the diagonals do not go!
PS: Sorry for my english
Like in image:
enter image description here
答案1
得分: 1
// 准备输出矩阵并用-1填充
int[][] outMatrix = prepareOut(inputArray.length, inputArray[0].length);
// 从初始坐标开始递归调用方法标记单元格
outMatrix = markCell(inputArray, outMatrix, line, column, 1);
// 打印输出矩阵
printOutput(outMatrix);
最相关的方法实现可能如下所示：
static int[][] markCell(int[][] arr, int[][] out, int y, int x, int move) {
int val = arr[y][x];
if (out[y][x] == 0) {
return out;
} else if (out[y][x] < 0) {
out[y][x] = move;
}
// 检查当前单元格上方的单元格（北）
if (y > 0 && out[y - 1][x] < 0) {
if (cellMarked(arr, out, y - 1, x, val, move)) {
out = markCell(arr, out, y -1, x, move + 1);
}
}
// 检查当前单元格下方的单元格（南）
if (y < arr.length - 1 && out[y + 1][x] < 0) {
if (cellMarked(arr, out, y + 1, x, val, move)) {
out = markCell(arr, out, y +1, x, move + 1);
}
}
// 检查当前单元格左侧的单元格（西）
if (x > 0 && out[y][x - 1] < 0) {
if (cellMarked(arr, out, y, x - 1, val, move)) {
out = markCell(arr, out, y, x - 1, move + 1);
}
}
// 检查当前单元格右侧的单元格（东）
if (x < arr[0].length - 1 && out[y][x + 1] < 0) {
if (cellMarked(arr, out, y, x +1, val, move)) {
out = markCell(arr, out, y, x + 1, move + 1);
}
}
return out;
}
static boolean cellMarked(int[][] arr, int[][] out, int y, int x, int val, int move) {
final boolean marked = arr[y][x] <= val;
out[y][x] = marked ? move : 0;
return marked;
}
在打印输出矩阵时，将剩余的-1替换为0：
static void printOutput(int[][] arr) {
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[i].length; j++) {
char c = arr[i][j] <= 0 ? '0' : '*';
System.out.print(c);
System.out.print('\t');
}
System.out.print('\n');
}
}
英文:
Three steps need to be done here:
// prepare output matrix and fill it with -1
int[][] outMatrix = prepareOut(inputArray.length, inputArray[0].length);
// call recursively method to mark cells starting from the initial coordinates
outMatrix = markCell(inputArray, outMatrix, line, column, 1);
// print output matrix
printOutput(outMatrix);
The most relevant method implementation may be like this:
static int[][] markCell(int[][] arr, int[][] out, int y, int x, int move) {
int val = arr[y][x];
if (out[y][x] == 0) {
return out;
} else if (out[y][x] < 0) {
out[y][x] = move;
}
// checking a cell above the current one (north)
if (y > 0 && out[y - 1][x] < 0) {
if (cellMarked(arr, out, y - 1, x, val, move)) {
out = markCell(arr, out, y -1, x, move + 1);
}
}
// checking a cell under the current one (south)
if (y < arr.length - 1 && out[y + 1][x] < 0) {
if (cellMarked(arr, out, y + 1, x, val, move)) {
out = markCell(arr, out, y +1, x, move + 1);
}
}
// checking a cell to the left of the current one (west)
if (x > 0 && out[y][x - 1] < 0) {
if (cellMarked(arr, out, y, x - 1, val, move)) {
out = markCell(arr, out, y, x - 1, move + 1);
}
}
// checking a cell to the right of the current one (east)
if (x < arr[0].length - 1 && out[y][x + 1] < 0) {
if (cellMarked(arr, out, y, x +1, val, move)) {
out = markCell(arr, out, y, x + 1, move + 1);
}
}
return out;
}
static boolean cellMarked(int[][] arr, int[][] out, int y, int x, int val, int move) {
final boolean marked = arr[y][x] <= val;
out[y][x] = marked ? move : 0;
return marked;
}
When printing the output matrix, you replace remaining -1 with 0:
static void printOutput(int[][] arr) {
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[i].length; j++) {
char c = arr[i][j] <= 0 ? '0' : '*';
System.out.print(c);
System.out.print('\t');
}
System.out.print('\n');
}
}
答案2
得分: 0
prepareOut可以像这样实现：
private static int[][] prepareOut(int rows, int cols) {
int [][] out = new int[rows][cols];
for(int[] row: out) {
Arrays.fill(row, -1);
}
return out;
}
英文:
prepareOut may be implemented like this:
private static int[][] prepareOut(int rows, int cols) {
int [][] out = new int[rows][cols];
for(int[] row: out) {
Arrays.fill(row, -1);
}
return out;
}
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