【发布时间】：2020-04-18 15:43:22
【问题描述】：
我们正在做以下练习：Snail。
为了能够完成程序我们想到了下面的例子：
给定数组：
[1, 2, 3, 1]
[4, 5, 6, 4]
[7, 8, 9, 7]
[7, 8, 9, 7]
1. Iterate over first row (i==0) j++
[0,0]=1, [0,1]=2, [0,2]=3, [0,3]=1
2. Traverse last column (j==array[i].length-1) i++
[1,3]=4, [2,3]=7, [3,3]=7
3. Iterate over last row (i==array.length-1) j--
[3,2]=9, [3,1]=8, [3,0]=7
4. Go up into first column, do not take first row (j==0) (0>i<array.length-1) i--
[2,0]=7, [1,0]=4
5. Iterate over second row, do not take the last column (i==1) (-1>j<array.length-2) j++
[1,1]=5, [1,2]=6
6. Get middle elements
[2,2]=9, [2,1]=8
我们编写了以下代码：
import java.util.*;
import java.util.stream.*;
public class Snail {
public static int[] snail /*????????????*/ (int[][] array) {
System.out.println("array: "+Arrays.deepToString(array));
/*
1. Iterate over first row (i==0) j++
2. Traverse last column (j==array[i].length-1) i++
3. Iterate over last row (i==array.length-1) j--
4. Go up into first column, do not take first row (j==0) (0>i<array.length-1) i--
5. Iterate over second row, do not take the last column (i==1) (-1>j<array.length-2) j++
6. Do it and finally get the middle elements (i==2) (0<j<array[i].length-1)
*/
List<Integer> firstRow = getRow(0,0,array[0].length,1,array);
System.out.println("firstRow: "+Arrays.toString(firstRow.toArray()));
List<Integer> lastColumn = getColumn(array[0].length-1,1,array[0].length,1,array);
System.out.println("lastColumn: "+Arrays.toString(lastColumn.toArray()));
List<Integer> lastRow = getRow(array.length-1,(array[array.length-1].length) - 2,-1,-1,array);
System.out.println("lastRow: "+Arrays.toString(lastRow.toArray()));
List<Integer> firstColumn = getColumn(0,array[0].length-2,0,-1,array);
System.out.println("firstColumn: "+Arrays.toString(firstColumn.toArray()));
List<Integer> middle = getRow(1,1,array[1].length-1,1,array);
System.out.println("middle: "+Arrays.toString(middle.toArray()));
List<Integer> middle2 = getRow(2,array[1].length-2,0,-1,array);
System.out.println("middle2: "+Arrays.toString(middle2.toArray()));
middle.addAll(middle2);
firstColumn.addAll(middle);
lastRow.addAll(firstColumn);
lastColumn.addAll(lastRow);
firstRow.addAll(lastColumn);
System.out.println("firstRow: "+Arrays.toString(firstRow.toArray()));
return firstRow.stream().mapToInt(i->i).toArray();
}
public static List<Integer> getRow(int row, int from, int to, int modifier, int[][] array){
List<Integer> result = new ArrayList<>();
for(int j=from; Math.max(j,to)>Math.min(j,to); j+=modifier){
System.out.println("j: "+j);
result.add(array[row][j]);
}
return result;
}
public static List<Integer> getColumn(int column, int from, int to, int modifier, int[][] array){
List<Integer> result = new ArrayList<>();
for(int j=from; Math.max(j,to)>Math.min(j,to); j+=modifier){
System.out.println("j: "+j);
result.add(array[j][column]);
}
return result;
}
}
如您所见，我们已按照步骤操作。但是，无论给定数组的大小，我们如何制定这个解决方案？
目前发布的代码通过了第一个测试，但没有通过第二个测试（所以我们需要一种方法来回答所有可能的数组大小的练习）：
import org.junit.Assert;
import org.junit.Test;
import org.junit.runners.JUnit4;
import java.util.Arrays;
import java.util.Random;
import static java.util.stream.Collectors.joining;
public class SnailTest {
@Test
public void SnailTest0() {
int[][] array
= {{1, 2, 3, 1},
{4, 5, 6, 4},
{7, 8, 9, 7},
{7, 8, 9, 7}};
int[] r = {1, 2, 3, 1, 4, 7, 7, 9, 8, 7, 7, 4, 5, 6, 9, 8};
test(array, r);
}
@Test
public void SnailTest1() {
int[][] array
= {{1, 2, 3},
{4, 5, 6},
{7, 8, 9}};
int[] r = {1, 2, 3, 6, 9, 8, 7, 4, 5};
test(array, r);
}
public String int2dToString(int[][] a) {
return Arrays.stream(a).map(row -> Arrays.toString(row)).collect(joining("\n"));
}
public void test(int[][] array, int[] result) {
String text = int2dToString(array) + " should be sorted to " + Arrays.toString(result);
System.out.println(text);
Assert.assertArrayEquals( result, Snail.snail(array));
}
}
我们如何改进这个算法来遍历一个从外部元素到内部元素的数组？你将如何修改这个算法以能够处理给定的任何数组大小？
【问题讨论】：
-
您的列表达式错误。而不是 'array[i].length-1' 你应该使用 'array[i][length-1]' 这将获得第 i 行最后一列的结果。