假设有一个1D数组test[]={1,2,3}
和2D数组arr1[3][5]={{1,2,5,4,3},{3,7,1,4,2},{2,9,7,8,3}}
。
输出所需的内容是:
test is the subset of row 0 of arr1
test is the subset of row 1 of arr1
test is not the subset of row 2 of arr1
这是我到目前为止已实施的代码:
class GFG {
public static void main(String args[]) {
int arr1[][] = { { 11, 1, 13, 3, 7 },
{ 11, 1, 17, 7, 3 },
{ 2, 5, 8, 9, 10 } };
int test[] = { 11, 3, 7, 1 };
int m = arr1.length; // rows
int n = test.length;
int o = arr1[0].length; // no. of elements in each row
System.out.println(o); // just for testing if it works
int i = 0;
int j = 0;
int k = 0;
for (i = 0; i < n; i++) {
for (j = 0; j < m && j != m; j++) {
for (k = 0; k < o; k++)
if (test[i] == arr1[j][k])
break;
if (k == o)
System.out.println("test[] is " + "not a subset of arr1 " + j + " row");
else
System.out.println("test[] is " + "subset of arr1 " + j + " row");
}
}
}
}
但是我从中获得的输出是:
在这里输出
我意识到这是I循环努力反复打印它,但在这种情况下,我并没有获得令人满意的输出。
在这里可以做什么?还是该问题有很多最佳的实现?欢迎任何建议。
您弄乱了周期的订单:您应该先迭代arr
。这个想法是ti使用标志: isSubset
,当连续发现某些元素contains
时,它会变成false,如果当前检查元素在一排。
可以进行改进(例如foreach循环和标记的休息),但我决定保持代码简单。
public class GFG {
public static void main(String args[]) {
int arr[][] = { { 11, 1, 13, 3, 7 },
{ 11, 1, 17, 7, 3 },
{ 2, 5, 8, 9, 10 } };
int test[] = { 11, 3, 7, 1 };
for (int i = 0; i < arr.length; i++) {
boolean isSubset = true;
for (int j = 0; j < test.length; j++) {
boolean contains = false;
for (int k = 0; k < arr[i].length; k++) {
if (test[j] == arr[i][k]) {
contains = true;
break;
}
}
if (!contains) {
isSubset = false;
break;
}
}
if (isSubset) {
System.out.println("test[] is " + "subset of arr " + i + " row");
} else {
System.out.println("test[] is " + "not a subset of arr " + i + " row");
}
}
}
}
尝试将解决方案分解为较小的方法,因此您的代码将更清晰,您可以更轻松地考虑正在发生的事情。这是一个可以完成的示例:
class GFG {
public static void main(String args[]) {
int arr1[][] = { { 11, 1, 13, 3, 7 },
{ 11, 1, 17, 7, 3 },
{ 2, 5, 8, 9, 10 } };
int test[] = { 11, 3, 7, 1 };
System.out.println(arr1[0].length); // just for testing if it works
for (int i = 0; i < arr1.length; i++) {
if (isSubset(test, arr1[i])) {
System.out.println("test[] is " + "subset of arr1 " + i + " row");
} else {
System.out.println("test[] is " + "not a subset of arr1 " + i + " row");
}
}
}
/* returns true if arr contains all elements of sub */
static boolean isSubset(int[] sub, int[] arr) {
for (int e : sub) {
if (!contains(e, arr)) return false;
}
return true;
}
/* returns true if arr contains elem */
static boolean contains(int elem, int[] arr) {
for (int e : arr) {
if (elem == e) return true;
}
return false;
}
}