2 回答

TA贡献1802条经验 获得超5个赞
你搞砸了循环的顺序:你应该先迭代arr。这个想法是 ti 使用 flags: isSubset,当在一行中找不到某个元素时,它变为 false,contains如果当前检查的元素在一行中,则变为 true。
可以进行一些改进(如 foreach 循环和标记中断),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");
}
}
}
}

TA贡献1876条经验 获得超5个赞
尝试将您的解决方案分解为更小的方法,这样您的代码会更清晰,您可以更容易地想到发生了什么。这是一个如何完成的示例:
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;
}
}
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